## Algebraic solution of a pair of linear equations in two variables by Cross-Multiplication

This 7 part series on **Pair of linear equations in two variables** complements Chapter 3 of NCERT Class 10 Maths.

In this fifth part the method of solving a pair of linear equations by **cross-multiplication** is covered.

*You may read the previous four parts of this seven part series using the following links.*

*NCERT solutions for class 10 maths linear equations 1 graphical representation,*

*NCERT solutions for class 10 maths linear equations 2 graphical solutions,*

*NCERT solutions for class 10 maths linear equations 3 algebraic solutions by substitution,*

*NCERT solutions for class 10 maths linear equations 4 algebraic solutions by elimination.*

**To elaborate further, in this fifth of the seven part series you'll,**

*Learn how to solve a pair of linear equations in two variables***algebraically by cross-multiplication method,***Learn the***concepts and mechanism**underlying the cross-multiplication method,**Learn how to remember the cross-multiplication fornula**for solution of a pair of linear equations easily, and*Learn how to apply substitution method, elimination method and cross-multiplication method***in solving the exercise problems.**

### Cross-multiplication method for solving a pair of linear equations in two variables

To apply the cross-multiplication method, **you need to remember first time a formula** for getting the values of $x$ and $y$ from a pair of linear equations in $x$ and $y$.

Before this, only very basic concepts were sufficient.

**Note:** *Over-dependence on formulas curbs your conceptual strength and the ability to think new.*

We'll reach the formula through already known concepts of elimination. It should help in remembering and using the formula.

In addition we'll point out the rules for remembering this popular formula easily.

Let's now learn how to solve two linear equations in two variables using cross-multiplication method by solving two problem examples.

#### Example problem 1.

Solve the following pair of linear equations algebraically.

$2x+3y-8=0$ and $4x+5y-7=0$.

#### Solution to example problem 1—algebraic solution of a pair of linear equations in two variables by cross-multiplication

The two equations are,

$2x+3y-8=0$ and,

$4x+5y-7=0$.

We'll assume that we don't know the cross-multiplication method. But we know that **for solving a pair of linear equations,**

**We have to eliminate one variable first.**- Next get the value of the variable remaining in the single linear equation.
- And then in the third step substitute the value of the variable thus obtained in any of the two equations to get the value of the second variable eliminated in the first step.

If you think for a moment about this situation you will appreciate the fact that,

You may decide to eliminate any of the two variables first—the two variables are equivalent in this respect.

In that case,

Why not take up the process of

eavaluating first variable by eliminating second variable,AND also evaluate the second variable by eliminating first variable TOGETHER!

This is the **basic motivation behind creation of this formula based cross-multiplication method.**

Let us see what happens if we * first eliminate $y$, get the value of $x$*, and then

*without substituting this value to get value of $y$,*

**again take up eliminating $x$ this time and get the value of $y$**.

**First: elimination of $y$ to get the value of $x$:**

$2x+3y-8=0$ and,

$4x+5y-7=0$.

To eliminate $y$ you have to multiply **first equation by coefficient $5$ of $y$ in the second equation**, and the *second equation by coefficient $3$ of $y$ in first equation.*

**Remember,** here you are going to eliminate equalized term of $y$ to get the value of $x$.

Results of the two multiplications are,

$5\times{2x}+5\times{3y}+5\times{(-8)}=0$, and

$3\times{4x}+3\times{5y}+3\times{(-7)}=0$.

Subtract the first equation above from the second (to keep $x$ positive). Result is,

$x(3\times{4}-5\times{2})=3\times{7}-5\times{8}$, terms in $y$ cancel out,

Or, $x=\displaystyle\frac{3\times{7}-5\times{8}}{3\times{4}-5\times{2}}$.

**Take note of this form of $x$ before getting the value as**,

$x=-\displaystyle\frac{19}{2}$.

**Second: elimination of $x$ to get the value of $y$:**

As decided, we won't substitute the value of $x$ in any of the equations to evaluate $y$. Instead we'll get the value of $y$ in the same way as we got the value of $x$, by eliminating $x$ instead of $y$.

Let's show you the two equations again for convenience. The two equations are,

$2x+3y-8=0$ and,

$4x+5y-7=0$.

To eliminate $x$, this time we have to multiply **first equation by** *coefficient $4$ of $x$* **in second equation** and **second equation by** *coefficient $2$ of $x$* **in first equation**. Results are,

$4\times{2x}+4\times{3y}+4\times{(-8)}=0$, and

$2\times{4x}+2\times{5y}+2\times{(-7)}=0$.

Subtract the second equation from the first (to make resultant coefficient of $y$ positive). Result is,

$y(4\times{3}-2\times{5})=4\times{8}-2\times{7}$,

Or, $y=\displaystyle\frac{4\times{8}-2\times{7}}{4\times{3}-2\times{5}}=9$.

Before looking at the value of $y$ at this point, let's examine this expression and the expression for $x$ more deeply.

**Denominators of variable values in cross-multiplication method**

Examine the **two denominators in the two expressions for $x$ and $y$**. The two are,

$3\times{4}-5\times{2}=4\times{3}-2\times{5}=2$. **Both are same.**

So we conclude that,

If we evaluate $x$ and $y$ by eliminating $y$ and $x$ respectively in two linear equations, the

denominators of both the values will be same.

It is time to introduce the **general form a linear equation**, $ax+by+c=0$.

Expressing both the original equations in this general form you get,

$a_1x+b_1y+c_1=0$, where $a_1=2$, $b_1=3$ and $c_1=-8$, and

$a_2x+b_2y+c_2=0$, where $a_2=4$, $b_2=5$ and $c_2=-7$.

In this form then the denominators of $x$ and $y$ will be same and will be equal to,

$a_2b_1-a_1b_2=4\times{3}-2\times{5}$.

**Essentially,** the two terms are formed by,

Multiplying coefficient of $x$ in second with coefficient of $y$ in the first, and then multiplying coefficient of $x$ in first by coefficient of $y$ in second equation.

This is where the name **cross-multiplication** comes from. Visually the situation is shown in the following graphic.

What about the numerators? As you can guess, to get the numerators also you have to cross-multiply, but this time multiply coefficients of the variables with the constant terms.

**Numerators of variable values in cross-multiplication method**

The** numerator of value of $x$** has been,

$3\times{7}-5\times{8}=-3\times{(-7)}+5\times{(-8)}=b_2c_1-b_1c_2$, replacing numeric values of coefficients by general form symbols.

The two terms are formed by multiplying coefficient of $y$ in second equation with constant term in first equation and multiplying coefficient of $y$ in first equation with constant term in second equation.

This also is cross-multiplication, but between coefficients of $y$ and constant terms. This is understandable, because if you remember—to evaluate $x$ you had eliminated $y$ by multiplying the equations with coefficients of $y$.

**Visually** formation of the numerator of $x$ is shown in the following graphic.

Similarly, the **numerator of value of $y$** has been,

$4\times{8}-2\times{7}=2\times{(-7)}-4\times{(-8)}=a_1c_2-a_2c_1$.

Again cross-multiplication, but this time between coefficients of $x$ with the constant terms.

**Visually**, formation of the numerator of $y$ is shown in the following graphic.

The values of $x$ and $y$ in general form can be expressed now as,

$x=\displaystyle\frac{b_2c_1-b_1c_2}{a_2b_1-a_1b_2}$,

Or. $\displaystyle\frac{x}{b_1c_2-b_2c_1}=\frac{1}{a_1b_2-a_2b_1}$, multiplying both sides by $-1$ to convert to a standard form where first factor of positive first term starts with parameter of equation 1.

$y=\displaystyle\frac{a_1c_2-a_2c_1}{a_2b_1-a_1b_2}$,

Or, $\displaystyle\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$, multiplying both sides by $-1$ to convert to a standard form where first factor of positive first term starts with parameter of equation 1.

RHS of the both the expression being equal, the two can be **combined into a single chained equation of three parts.**

**Combining values of two variables into a chained equation in a systematic form**

So combining the two expression above,

$\displaystyle\frac{x}{b_1c_2-b_2c_1}=\displaystyle\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$.

The denominators of the three equal components are formed by cross-multiplying two pairs of parameters of the two equations in a **systematic standard scheme** that is easy to remember following a few rules.

The **composite cross-multiplication solution scheme** is shown below in the following graphic,

**RULES to remember the formula for solving two linear equations in two variables by cross-multiplication**

The two equations in $x$ and $y$ are,

$a_1x+b_1y+c_1=0$, and

$a_2x+b_2y+c_2=0$.

Solution values of $x$ and $y$ are given by,

$\displaystyle\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$.

The equal denominator of the two values is in the third part of the chained equation.

**Rules to remember** this form of chained equation is straightforward:

- The three numerators will be in sequence of $x$, $y$ and $1$.
**Each of the denominators**will be formed by**cross-multiplying two pairs of parameter**values in the form: $(p_1q_2-p_2q_1)$.- The first, second and third denominator equivalents of
**first factor $p_1$ of the positive first term**will be: $b_1$, $c_1$ and $a_1$. - The sequence of parameter operation will be: $b⇒c⇒a⇒b$.
- Each parameter, say $a_1$ in a term will be
**cross-multiplied****with**the**next in sequence parameter of the other equation**, in this case with $b_2$.

If you remember the general form of the chained equation and the first factor of the first term of each denominator as, $b_1$, $c_1$ and $a_1$, the other factors and terms can easily be enumerated.

You should form your own rules for quick and correct recall of this popular formula.

**Solution of our problem example is** $x=-\displaystyle\frac{19}{2}$, $y=9$.

**Verify the solution** by substituting these two values in both the equations.

The result of substitution in the first equation is,

$2x+3y=8$,

Or, $-19+27=8$,

Or, $8=8$, the equation is satisfied.

Similarly substituting in the second equation you get,

$4x+5y=7$,

Or, $-38+45=7$,

Or, $7=7$.

The second equation is also satisfied. So **you are sure that the solution is correct.**

#### Example Problem 2.

The cost of 5 oranges and 3 apples is Rs.35 and the cost of 2 oranges and 4 apples is Rs.28. Find the cost of an orange and that of an apple.

#### Solution to Example Problem 2—solution of a pair linear equations algebraically by Cross-multiplication

In this example, we'll solve the two equations by **the method of cross-multiplication.**

Let's assume cost of an orange and that of an apple are, $x$ and $y$ respectively in rupees.

By the first statement the total cost of the purchase is represented by the linear equation,

$5x+3y=35$,

Or, $5x+3y-35=0$, converting in standard form $ax+by+c=0$ for applying cross-multiplication method.

Similarly by the second statement the total cost of purchase is represented by the second equation,

$2x+4y=28$,

Or, $x+2y-14=0$, **eliminating common factor 2** and expressing in the standard form of $ax+by+c=0$.

**It is always good to eliminate a common factor (the largest) from an equation to save calculation time.**

If you have memorized the cross-multiplication formula by now, it is easy to write down the chained $x$-$y$ value equation.

Instead, let us form the cross-multiplication parameter graphic for the problem as shown below. In this problem, $a_1=5$, $b_1=3$, $c_1=-35$, $a_2=1$, $b_2=2$ and $c_2=-14$.

Take your time to examine the graphic. It is self-explanatory and includes the solution also.

**Recommendation:** Always **follow the** **cross-multiplication arrow diagram scheme of representation of relationships between the parameter values** in **forming the product terms** **STRICTLY**. You'll never be wrong then in **forming the resultant chained equation from the cross-multiplication scheme.**

The **chained equation for values of $x$ and $y$** as shown in the graphic above is,

$\displaystyle\frac{x}{3\times{(-14)}-2\times{(-35)}}$

$=\displaystyle\frac{y}{(-35)\times{1}-(-14)\times{5}}$

$=\displaystyle\frac{1}{5\times{2}-1\times{3}}$,

Or, $\displaystyle\frac{x}{28}=\frac{y}{35}=\frac{1}{7}$,

So, $x=4$ and $y=5$.

**Answer:** Cost of an orange is Rs.4 and that of an apple is Rs.5.

**Verification: Substitute these cost values in the problem statements directly, and NOT IN THE EQUATIONS** because *forming the equations may itself be erroneous.*

Result will be, first total cost: $(5\times{4}+3\times{5})=20+15=35$. Correct.

Second total cost: $(2\times{4}+4\times{5})=8+20=28$. Correct again. Solution verified successfully.

We'll now solve the problems in the exercise 3.5.

### Problems to solve as in Exercise

#### Problem 1.

Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using cross-multiplication method.

- $x-3y-3=0$ and $3x-9y-2=0$
- $2x+y=5$ and $3x+2y=8$
- $3x-5y=20$ and $6x-10y=40$
- $x-3y-7=0$ and $3x-3y-15=0$

#### Problem 2.

- For which values of $a$ and $b$ does the following pair of linear equations have an infinite number of solutions?

$\hspace{12mm} 2x+3y=7$

$\hspace{12mm}(a-b)x+(a+b)y=3a+b-2$

- For which value of $k$ will the following pair of linear equations have no solution?

$\hspace{12mm}3x+y=1$

$\hspace{12mm}(2k-1)x+(k-1)y=2k+1$

#### Problem 3.

Solve the following pair of linear equations by substitution and cross-multiplication methods.

$\hspace{12mm}8x+5y=9$ and $3x+2y=4$

#### Problem 4.

Form the linear equations in the following problems and find their solutions (if they exist) by any algebraic method.

- A part of monthly hostel charges are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes for 20 days she has to pay Rs.1000 as hostel charges whereas a student B, who takes food for 26 days pays Rs.1180 as hostel charges. Find the fixed charge and the cost of food per day.
- A fraction becomes $\displaystyle\frac{1}{3}$ when 1 is subtracted from the numerator and it becomes $\displaystyle\frac{1}{4}$ when 8 is added to the denominator. Find the fraction.
- Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
- Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the cars?
- The area of a rectangle gets reduced by 9 square units if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimension of the rectangle.

### Solution to the problems

#### Problem 1.i.

Which of the following pair of linear equation has unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using cross-multiplication method.

- $x-3y-3=0$ and $3x-9y-2=0$

#### Solution to Problem 1.i: Check for unique solution

The two equations are,

$x-3y-3=0$ and

$3x-9y-2=0$

The ratios of coefficients of $x$ and $y$ are equal to each other but not equal to the ratio of the constants.

The ratios are,

$\displaystyle\frac{a_1}{a_2}=\frac{1}{3}$

$\displaystyle\frac{b_1}{b_2}=\frac{-3}{-9}=\frac{1}{3}=\frac{a_1}{a_2}$.

$\displaystyle\frac{c_1}{c_2}=\frac{-3}{-2} \neq \displaystyle\frac{a_1}{a_2}$.

The straight lines representing the two linear equations are parallel and would have no solution.

**Answer:** No solution.

#### Problem 1.ii.

Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using cross-multiplication method.

- $2x+y=5$ and $3x+2y=8$

#### Solution to Problem 1.ii: Check for unique solution and solve by cross-multiplication

The two equations are,

$2x+y=5$,

Or, $2x+y-5=0$ and,

$3x+2y=8$,

Or, $3x+2y-8=0$.

The ratios of the coefficients of the two variables and the constants are,

$\displaystyle\frac{a_1}{a_2}=\frac{2}{3}$

$\displaystyle\frac{b_1}{b_2}=\frac{1}{2}$

$\displaystyle\frac{c_1}{c_2}=\frac{-5}{-8}=\frac{5}{8}$.

All three ratios being unequal the two equations would have a unique solution.

Fitting the two equations in their general form you get,

$a_1x+b_1y+c_1=0$, where $a_1=2$, $b_1=1$ and $c_1=-5$, and

$a_2x+b_2y+c_2=0$, where $a_2=3$, $b_2=2$, and $c_2=-8$.

The chained equation giving values of $x$ and $y$ in the solution is,

$\displaystyle\frac{x}{b_1c_2-b_2c_1}=\displaystyle\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$,

Or, $\displaystyle\frac{x}{1\times{(-8)}-2\times{(-5)}}$

$=\displaystyle\frac{y}{(-5)\times{3}-(-8)\times{2}}$

$=\displaystyle\frac{1}{2\times{2}-3\times{1}}$,

Or, $\displaystyle\frac{x}{2}=\frac{y}{1}=\frac{1}{1}$,

Or, $x=2$, and $y=1$.

**Answer:** $x=2$, and $y=1$.

**Verification:** Substitute the values of $x$ and $y$ in the two equations to get respectively,

$5=5$, and

$8=8$.

The values satisfy both the equations and so is the correct solution. Verification increases your confidence on the solution.

The **cross-multiplication scheme** is shown in the following graphic.

#### Problem 1.iii.

Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using cross-multiplication method.

- $3x-5y=20$ and $6x-10y=40$

#### Solution to Problem 1.iii: Check for unique solution

The two equations are,

$3x-5y=20$,

Or, $3x-5y-20=0$ and,

$6x-10y=40$,

Or, $6x-10y-40=0$.

The ratios of the coefficients of the two variables and the constants are,

$\displaystyle\frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2}$

$\displaystyle\frac{b_1}{b_2}=\frac{-5}{-10}=\frac{1}{2}$

$\displaystyle\frac{c_1}{c_2}=\frac{-20}{-40}=\frac{1}{2}$.

All three ratios being equal, the equations are equivalent and the corresponding lines are coincident with infinitely many solutions.

**Answer:** Infinitely many solutions.

#### Problem 1.iv.

- $x-3y-7=0$ and $3x-3y-15=0$

#### Solution to Problem 1.iv: Check for unique solution and solving by cross-multiplication

The equations are,

$x-3y-7=0$ and

$3x-3y-15=0$.

The ratios of the coefficients of the two variables and the constants are,

$\displaystyle\frac{a_1}{a_2}=\frac{1}{3}$

$\displaystyle\frac{b_1}{b_2}=\frac{-3}{-3}=1$

$\displaystyle\frac{c_1}{c_2}=\frac{-7}{-15}=\frac{7}{15}$.

All three ratios being unequal, the two equations would have a unique solution.

Fitting the two equations in their general form you get,

$a_1x+b_1y+c_1=0$, where $a_1=1$, $b_1=-3$ and $c_1=-7$, and

$a_2x+b_2y+c_2=0$, where $a_2=3$, $b_2=-3$, and $c_2=-15$.

The chained equation giving values of $x$ and $y$ in the solution is,

$\displaystyle\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$,

Or, $\displaystyle\frac{x}{(-3)\times{(-15)}-(-3)\times{(-7)}}$

$=\displaystyle\frac{y}{(-7)\times{3}-(-15)\times{1}}$

$=\displaystyle\frac{1}{1\times{(-3)}-3\times{(-3)}}$,

Or, $\displaystyle\frac{x}{24}=\frac{y}{-6}=\frac{1}{6}$,

Or, $x=4$, and $y=-1$.

**Answer:** $x=4$ and $y=-1$.

**Verification:** Substitute these two values to the first and second equation respectively to get,

$7-7=0$, and

$15-15=0$.

Solution verified.

The **cross multiplication scheme** is shown in the following graphic.

#### Problem 2.i.

- For which values of $a$ and $b$ does the following pair of linear equations have an infinite number of solutions?

$\hspace{12mm} 2x+3y=7$

$\hspace{12mm}(a-b)x+(a+b)y=3a+b-2$

#### Solution to Problem 2.i: Conditions for infinitely many solutions

The two equations are,

$2x+3y=7$, and

$(a-b)x+(a+b)y=3a+b-2$

The ratios of the coefficients of the two variables and the constants are,

$\displaystyle\frac{a_1}{a_2}=\frac{2}{(a-b)}$

$\displaystyle\frac{b_1}{b_2}=\frac{3}{(a+b)}$

$\displaystyle\frac{c_1}{c_2}=\frac{7}{(3a+b-2)}$.

For the two equations to be equivalent and infinitely many solutions, the three ratios must be equal. That is,

$\displaystyle\frac{2}{(a-b)}=\frac{3}{(a+b)}=\frac{7}{(3a+b-2)}$.

Taking the equality of the first two ratios,

$\displaystyle\frac{2}{(a-b)}=\frac{3}{(a+b)}$,

Or, $3(a-b)=2(a+b)$,

Or, $a=5b$.

Now take the equality of the second and the third ratios,

$\displaystyle\frac{3}{(a+b)}=\frac{7}{(3a+b-2)}$,

Or, $3(3a+b-2)=7(a+b)$,

Or, $2a-4b=6$.

Subsitute expression for $a$,

$10b-4b=6$.

Or, $b=1$.

So, $a=5$.

**Answer:** $a=5$, $b=1$.

**Verification:** With $x=5$ and $y=1$, the three ratios become,

$\displaystyle\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}$

$\displaystyle\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}$

$\displaystyle\frac{c_1}{c_2}=\frac{7}{14}=\frac{1}{2}$.

All three ratios of parameter pairs being equal the equations are equivalent with infinitely many solutions.

#### Problem 2.ii.

- For which value of $k$ will the following pair of linear equations have no solution?

$\hspace{12mm}3x+y=1$

$\hspace{12mm}(2k-1)x+(k-1)y=2k+1$

#### Solution to Problem 2.ii: Condition for no solution

The two equations are,

$3x+y=1$,

Or, $3x+y-1=0$,

$(2k-1)x+(k-1)y=2k+1$,

Or, $(2k-1)x+(k-1)y-(2k+1)=0$

The two equations to have no solution, the ratio of coefficients of $x$ and $y$ must be equal (for the slopes to be equal) so that the lines are parallel. Also ratio of constants terms must be unequal to the other two equal ratios (otherwise there would be infinitely many solutions).

The three ratios of the coefficients of two variables and pair of constant terms are,

$\displaystyle\frac{a_1}{a_2}=\frac{3}{(2k-1)}$

$\displaystyle\frac{b_1}{b_2}=\frac{1}{(k-1)}$, and

$\displaystyle\frac{c_1}{c_2}=\frac{-1}{-(2k+1)}=\frac{1}{2k+1}$.

For the lines corresponding to the equations to be parallel the first two ratios must be equal, that is,

$\displaystyle\frac{3}{(2k-1)}=\frac{1}{(k-1)}$,

Or, $3(k-1)=2k-1$,

Or, $k=2$.

**Answer:** $k=2$ for no solution.

**Verification:** With $k=2$, the three ratios are,

$\displaystyle\frac{a_1}{a_2}=\frac{3}{(2k-1)}=\frac{3}{3}=1$,

$\displaystyle\frac{b_1}{b_2}=\frac{1}{(k-1)}=\frac{1}{1}=1$, but

$\displaystyle\frac{c_1}{c_2}=\frac{1}{2k+1}=\frac{1}{5}$, unequal to the other two equal ratios. Solution verified.

#### Problem 3.

Solve the following pair of linear equations by substitution and cross-multiplication methods.

$\hspace{12mm}8x+5y=9$ and $3x+2y=4$.

#### Solution to Problem 3 by Substitution method

The two equations are,

$8x+5y=9$ and,

$3x+2y=4$.

Express $y$ in terms of $x$ from the second equation as it is simpler. Result is,

$y=\displaystyle\frac{4-3x}{2}$.

Substitute this expression for $y$ in the first equation. Result is,

$8x+5\times{\displaystyle\frac{4-3x}{2}}=9$

Or, $x=18-20=-2$.

Substitute this value of $x$ in the expression for $y$. Result is,

$y=\displaystyle\frac{4-3\times{(-2)}}{2}=\frac{10}{2}=5$.

**Answer:** $x=-2$, $y=5$.

**Verification:** Substituting the two values in the first and second equation respectively you would get,

$-16+25=9-9$, and

$-6+10=4=4$. Solution verified.

#### Solution to Problem 3 by cross-multiplication method

The two equations are,

$8x+5y=9$.

Or, $8x+5y-9=0$ and,

$3x+2y=4$,

Or, $3x+2y-4=0$.

Fitting the two equations in their general form you get,

$a_1x+b_1y+c_1=0$, where $a_1=8$, $b_1=5$ and $c_1=-9$, and

$a_2x+b_2y+c_2=0$, where $a_2=3$, $b_2=2$, and $c_2=-4$.

The chained equation giving values of $x$ and $y$ in the solution is,

$\displaystyle\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$,

Or, $\displaystyle\frac{x}{5\times{(-4)}-2\times{(-9)}}=\frac{y}{(-9)\times{3}-(-4)\times{8}}=\frac{1}{8\times{2}-3\times{5}}$,

Or, $\displaystyle\frac{x}{-2}=\frac{y}{5}=\frac{1}{1}$,

Or, $x=-2$, and $y=5$.

The same answer you have got by substitution method.

The **parameter cross-multiplication scheme** and **chained equation formation** is shown in the **composite graphic** below.

**Note:** Judge which method is more easy and quick for you.

#### Solution to Problem 4.i.

Form the linear equations in the following problems and find their solutions (if they exist) by any algebraic method.

- A part of monthly hostel charges are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs.1000 as hostel charges whereas a student B, who takes food for 26 days pays Rs.1180 as hostel charges. Find the fixed charge and the cost of food per day.

Let fixed monthly hostel fees be $x$ and the per day food cost be $y$ in rupees.

For student A and B, total costs are represented respectively by,

$x+20y=1000$, and

$x+26y=1180$.

Subtract first equation from second to eliminate $x$ and get,

$6y=180$,

$y=30$.

Substitute value of $y$ in first equation,

$x+600=1000$,

Or, $x=400$.

**Answer:** Fixed charge is Rs.400 and per day food charge is Rs.30.

This is the easiest and quickest method, for us.

#### Solution to Problem 4.ii

Form the linear equations in the following problems and find their solutions (if they exist) by any algebraic method.

- A fraction becomes $\displaystyle\frac{1}{3}$ when 1 is subtracted from the numerator and it becomes $\displaystyle\frac{1}{4}$ when 8 is added to the denominator. Find the fraction.

Let the numerator of the fraction be $x$ and the denominator $y$.

By the first condition then,

$\displaystyle\frac{x-1}{y}=\frac{1}{3}$

Or, $3x-y=3$.

By the second condition,

$\displaystyle\frac{x}{y+8}=\frac{1}{4}$,

Or, $4x-y=8$

Subtracting first from second equation,

$x=5$.

Substituting this value in first equation,

$y=15-3=12$.

**Answer:** Fraction is, $\displaystyle\frac{5}{12}$.

Verify the correctness of the answer yourself by applying the given conditions on the answer value of the fraction.

#### Solution to Problem 4.iii

- Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

Let the number of correct and wrong answers given by Yash be $x$ and $y$.

Area of a rectangle is given by the product of its length and breadth.

By the first statement then,

$3x-y=40$.

And by the second statement,

$4x-2y=50$.

Multiply the first equation by 2 to get,

$6x-2y=80$.

Subtract from this equation, the second equation formed above. Result is,

$2x=30$,

Or, $x=15$.

Substitute this value in first equation,

$45-y=40$,

Or, $y=5$.

**Answer:** Yash answered 15 questions correctly and 5 questions wrongly. So in total Yash answered 20 questions in the test.

#### Solution to Problem 4.iv.

- Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the cars?

Let the speed of the car starting from A be $x$ and that of car starting from B be $y$ both in km/hour.

In the first instance, when the cars travel in the same direction, the relative speed of the car moving faster, say car starting from A, is difference between the two speeds.

With this speed of $(x-y)$, the car A will move towards the second car virtually and effectively stationary at B.

**Note:** both are actually moving, but effectively first car is moving towards the second at a relative speed obtained by subtracting the speed of second car from that of the first car. That's why the second car is assumed standing effectively still when the first car moves towards it at the relative speed. At this speed of $(x-y)$ the faster car will traverse the initial separation of 100 km.

So in 5 hours Car A will cover the initial distance gap of 100 km and will catch up with car B.

This is represented by the equation,

$5(x-y)=100$,

Or, $x-y=20$.

Again for the second situation when the cars move towards each other, the distance of 100 km is covered at relative speed of $(x+y)$ in 1 hour. This is represented by the equation,

$x+y=100$.

Add the two and divide by 2 to get,

$x=60$.

Substitute this value in second equation to get value of second car as,

$y=40$.

**Answer:** The speeds of the two cars are 60 km per hour and 40 km per hour.

#### Solution to Problem 4.v.

- The area of a rectangle gets reduced by 9 square units if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimension of the rectangle.

Let length and breadth of the rectangle is $x$ and $y$ respectively both in length units.

By the first statement then,

$(x-5)(y+3)=xy-9$,

Or, $xy +3x-5y -15=xy-9$,

Or, $3x-5y=6$, a linear equation.

Similarly by the second statement,

$(x+3)(y+2)=xy+67$,

Or, $xy+2x+3y +6=xy+67$,

Or, $2x+3y=61$, a second linear equation.

Multiply the first equation by $\displaystyle\frac{3}{5}$ to transform it to,

$1.8x-3y=3.6$.

Add it to the second equation,

$3.8x=64.6$,

Or, $38x=646$,

Or, $2x=34$,

Or, $x=17$.

Substitute it to first equation $3x-5y=6$,

$51-5y=6$,

Or, $5y=45$,

Or, $y=9$.

**Answer:** Length is 17 units and breadth 9 units.

Verify yourself by using these values on the problem conditions. Do not substitute in the equations (equations might have been wrongly formed).

Note:Observe that in spite of the coefficients being such that multiplication and changing both equations would have been required, we have changed only one equation by usingfraction multiplier. To us this is the fastest and easiest method of solution of two linear equations in two variables.You may consider this to be an

additional method in your repertoire.You may call this method as—Solving two linear equations in two variables by variable elimination using FRACTION MULTIPLIER.

Let's quickly tell you formally this fourth method of solving a pair of linear equations.

### Fraction multiplier method of solving two linear equations in two variables

**Problem**

Solve the linear equations, $4x+15y=72$ and $6x+7y=46$.

#### Solution

**First decision:** Decide $x$ to be eliminated because the fraction multiplier of the first equation $\displaystyle\frac{3}{2}$ is small in both numerator and denominator.

**First action:** Multiply first equation by this fraction multiplier to get,

$6x+\displaystyle\frac{45}{2}y=108$

**Elimination of $x$:** Subtract second from first. Result is,

$\displaystyle\frac{31}{2}y=62$.

Or, $y=4$, and

$x=3$.

**Note:** This is not a prescribed method in text. So if you use the method, use it after taking required steps.

In the sixth part next, we'll learn how to solve special kinds of a pair of equations that do not seem to be linear.

### NCERT Solutions for Class 10 Maths

#### Chapter 1: Real Numbers

**NCERT Solutions for Class 10 Maths on Real numbers part 1, Euclid’s division lemma puzzle solutions**

#### Chapter 2: Polynomials

#### Chapter 3: Linear Equations

**NCERT solutions for class 10 maths Chapter 3 Linear equations 7 Problem Collection**

**NCERT solutions for class 10 maths Chapter 3 Linear equations 6 Reducing non-linear to linear form**

**NCERT solutions for class 10 maths Chapter 3 Linear equations 5 Algebraic solution by Cross Multiplication**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 4 Algebraic solution by Elimination**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 3 Algebraic solution by Substitution**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 2 Graphical solutions**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 1 Graphical representation.**

#### Chapter 4: Quadratic equations

**NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 1 What are quadratic equations**

**NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 2 Solving by factorization**

**NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 3 Solution by Completing the square**

#### Chapter 8: Introduction to Trigonometry, Concepts and solutions to exercise problems

*NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 1 Trigonometric Ratios*

*NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 4 Trigonometric identities*

#### Chapter 8: Introduction to Trigonometry, only solutions to selected problems

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 6**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 5**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 4**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 3**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 2**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 1**