Pair of Linear Equations in Two Variables

1.0Simultaneous linear equations in two variables

A pair of linear equations in two variables is said to form a system of simultaneous linear equations. General form : ${\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0$ and ${\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2=0$, where ${\mathrm{a}}_1,{\mathrm{a}}_2,{\mathrm{b}}_1,{\mathrm{b}}_2,{\mathrm{c}}_1$ and ${\mathrm{c}}_2$ are real non zero numbers; $a_1^2+b_1^2\ne 0$ and $a_2^2+b_2^2\ne 0$ and $x$, $y$ are variables. Here is an example of a system of simultaneous linear equations. $x-2y=3,2x+5y=5$

• $x=2\&y=-1$ is a solution a linear equation $3x+2y=4$, because values of $x\&y$ satisfies the equation. i.e. LHS $=3(2)+2(-1)$ = 6-2 $=4$ = RHS

Solution of simultaneous linear equations in two variables

A pair of values of the variables $x$ and $y$ satisfying each one of the equations in a given system of two simultaneous linear equations in $x$ and $y$ is called a solution of the system. E.g. $x=2,y=3$ is a solution of the system of simultaneous linear equations. $2x+y=7$ ..(1) $3x+2y=12$ ..(2) Put $x=2,y=3$ in L.H.S of equation (1), we get L.H.S $\Rightarrow 2\times 2+3=7=$ R.H.S Put $\mathrm{x}=2,\mathrm{y}=3$ in L.H.S of equation (2), we get L.H.S $=3\times 2+2\times 3=12=$ R.H.S The value of $x=2,y=3$ satisfy both equations (1) and (2).

2.0Investigating graphs of a system of equations

Recall from class IX that the geometrical representation of a linear equation in two variables is a straight line. Each solution ( $\mathrm{x},\mathrm{y}$ ) of a linear equation in two variables, $ax+by+c=0$, corresponds to a point on the line representing the equations, and vice versa. As such, there are infinitely many solutions of a linear equation in 2 variables. Since a linear equation in two variables represents a straight line. Therefore, a pair of linear equations in two variables will be represented by two straight lines, both to be considered together. You know that for given two lines in a plane, only one of the following three possibilities can happen: (i) The two lines intersect at one point. (ii) The two lines are parallel. (iii) The two lines are coincident lines.

Thus, the graphical representation of a pair of simultaneous linear equations in two variables will be in one of the forms. (see figure below)

Building Concepts 1 In one week a music store sold 7 violins for a total of ₹ $1500$. Two different types of violins were sold. One type cost ₹ 200 and the other type cost ₹ 300 . Represent this situation algebraically and graphically. Explanation:

points are $(0,7),(7,0),(3,4)$ $200\mathrm{x}+300\mathrm{y}=1500$ $\Rightarrow 2\mathrm{x}+3\mathrm{y}=1500$ $\Rightarrow y=\frac{15-2x}{3}$

points are $(0,5),(3,3),(-3,7)$ The graphical representation is shown in the figure which shows intersecting lines.

Building Concepts 2 Two railway tracks are represented by the equations $x+2y-6=0$ and $x+2y-4=0$. Represent this situation graphically. Explanation $$\begin{gathered} x+2y-6=0 \\ x+2y-4=0 \\ x+2y-6=0 \end{gathered}$$ $\Rightarrow y=\frac{6-x}{2}$

Points are $(0,3),(6,0),(2,2)$ $x+2y-4=0$ $\Rightarrow y=\frac{4-x}{2}$ Points are $(0,2),(4,0),(2,1)$

The graphical representation is shown in the figure which shows parallel lines.

Building Concepts 3 Lorea is moving along the path $\mathrm{x}+\mathrm{y}=6$ and Johana is moving along the path $2\mathrm{x}+2\mathrm{y}=12$. Represent this situation graphically. Explanation: $$\begin{gathered} x+y=6 \\ \Rightarrow y=6-x \\ 2\mathrm{x}+2\mathrm{y}=12 \\ \Rightarrow \mathrm{y}=\frac{12-2\mathrm{x}}{2} \end{gathered}$$

• At least 3 solutions are required to draw a graph of a linear equation in two variables.

3.0Solving system of equations by graphing

A system of two linear equations can be solved graphically, by graphing both equations in the same co-ordinate plane. Every point on the line of an equation is a solution of that equation.

The point at which the two lines cross lies on both lines and so it is the solution of both equations, it is also known as unique solution.

If the two lines coincide, they have infinitely many common points, and hence the system has infinitely many solutions.

Finally, if the lines represented by the equations are parallel, they do not have a common point and so the system has no solution.

Building Concepts 4 Solve the following system of equations by graphing. (i) $x+2y-3=0,4x+3y=2$ (ii) $3x+y=1,2y=2-6x$ (iii) $2x-y=2,2y-4x=2$ Solution:

(i) $x+2y-3=0$ $\Rightarrow y=\frac{3-x}{2}$

Points are $(1,1),(3,0),(-3,3)$

Points are $(2,-2),(-1,2),(5,-6)$ From the graph, (see fig.) we see that the two lines intersect at a point ( $-1,2$ ). So unique solution of the pair of linear equations is $x=-1,y=2$. (ii) $3x+y=1$ $\Rightarrow y=1-3x$

Points are $(0,1),(1,-2),(2,-5)$ $2y=2-6x$ $\Rightarrow y=\frac{2-6x}{2}$

Points are $(-1,4),(1,-2),(-2,7)$ The graph is shown in figure. The two equations have the same graph. Thus, system has infinite number of solutions.

(iii) $2\mathrm{x}-\mathrm{y}=2\Rightarrow \mathrm{y}=2\mathrm{x}-2$

Points are $(0,-2),(1,0),(2,2)$ $2y-4x=2\Rightarrow y=\frac{4x+2}{2}$

Points are $(0,1),(1,3),(-1,-1)$ The graph is shown alongside (see fig.). The graph of the system consists of two parallel lines. So, it has no solution.

• Plot three points for drawing a graph of linear equation in two variables. If these three points do not lie on same line that means at least one of the points (solution) is not correct.

4.0Types of System of Equation

Consistent system: A system of equations with at least one solution is called a consistent system (Unique and infinite many solutions).

Inconsistent system: A system of equations with no solution is called an inconsistent system (No solutions).

Dependent system: A system of equations with an infinite number of solutions is said to be dependent (Infinite many solutions).

Independent system: A system of equations with only one solution is said to be independent (Unique solutions).

Algebraic conditions for consistency/inconsistency of the system

• Linear equation in two variables is written in standard form as ${\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0$

Numerical Ability 1 In each of the following pairs of equations, determine whether the system has a unique solution, no solution or infinitely many solutions : (i) $2x+5y=17$ $5x+3y=14$ (ii) $x-3y-3=0$ $3x-9y-2=0$ (iii) $4x+6y=7$ $6x+9y=10.5$ Solution: The given pair of equations can be rewritten as $$\begin{gathered} \text{ (i) }2\mathrm{x}+5\mathrm{y}-17=0 \\ 5\mathrm{x}+3\mathrm{y}-14=0 \\ \text{ Here }{\mathrm{a}}_1=2,{\mathrm{b}}_1=5\text{ and }{\mathrm{c}}_1=-17 \\ {\mathrm{a}}_2=5,{\mathrm{b}}_2=3\text{ and }{\mathrm{c}}_2=-14 \\ \therefore \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{2}{5},\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{5}{3} \\ \text{ Clearly }\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}\ne \frac{{\mathrm{b}}_1}{{\mathrm{b}}_2} \end{gathered}$$ $\therefore$ The given pair of equations has a unique solution. (ii) The given pair of equations is $x-3y-3=0$ $3x-9y-2=0$ Here ${\mathrm{a}}_1=1,{\mathrm{b}}_1=-3$ and ${\mathrm{c}}_1=-3$ $a_2=3,b_2=-9$ and $c_2=-2$ $\therefore \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{1}{3},\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{-3}{-9}=\frac{1}{3}$ and $\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}=\frac{-3}{-2}=\frac{3}{2}$ $\therefore$ Clearly $\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}\ne \frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}$. $\therefore$ The given pair of equations has no solution. (iii) The given pair of equations can be rewritten as $4x+6y-7=0$ $6x+9y-10.5=0$ Here ${\mathrm{a}}_1=4,{\mathrm{b}}_1=6$ and ${\mathrm{c}}_1=-7$ $a_2=6,b_2=9$ and $c_2=-10.5$ $\therefore \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{4}{6}=\frac{2}{3},\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{6}{9}=\frac{2}{3}$ and $\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}=\frac{-7}{-10.5}=\frac{2}{3}$ $\therefore \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}$. $\therefore$ The given pair of equations has infinitely many solutions.

Numerical Ability 2 For what value of $k$, the system of equations $x+2y=5,3x+ky+15=0$ has (i) A unique solution (ii) No solution? Solution: We have, $x+2y=5$ $\Rightarrow \mathrm{x}+2\mathrm{y}-5=0$ and $3\mathrm{x}+\mathrm{ky}+15=0$. (i) The required condition for unique solution is: $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$ $\therefore \frac{1}{3}\ne \frac{2}{\mathrm{k}}\Rightarrow \mathrm{k}\ne 6$ Hence, for all real values of $k$ except 6 , the given system of equations will have a unique solution. (ii) The required condition for no solution is: $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$ $\therefore \frac{1}{3}=\frac{2}{\mathrm{k}}\ne \frac{-5}{15}$ $\Rightarrow \frac{1}{3}=\frac{2}{\mathrm{k}}$ and $\frac{2}{\mathrm{k}}\ne \frac{-5}{15}$ $\Rightarrow \mathrm{k}=6$ and $\frac{2}{\mathrm{k}}\ne \frac{-1}{3}$ $\Rightarrow \mathrm{k}=6$ and $\mathrm{k}\ne -6$ Hence the given system of equations will have no solution when $\mathrm{k}=6$.

Numerical Ability 3 Find the value of $k$ for which the system of equations $4x+5y=0,kx+10y=0$ has infinitely many solutions. Solution: The given system is of the form $a_{1}x+b_{1}y=0,a_{2}x+b_{2}y=0$ ${\mathrm{a}}_1=4,{\mathrm{a}}_2=\mathrm{k},{\mathrm{b}}_1=5,{\mathrm{b}}_2=10$ If the system has infinitely many solutions, $\frac{a_1}{a_2}=\frac{b_1}{b_2}$ $\therefore \frac{4}{\mathrm{k}}=\frac{5}{10}$ $\Rightarrow \mathrm{k}=8$

• For finding $\frac{a_1}{a_2},\frac{b_1}{b_2}\&\frac{c_1}{c_2}$, take coefficients $a_1,b_1,c_1,a_2,b_2\&c_2$ with signs before them in the equations.
• If the constant terms are zero in a simultaneous system of equations, the system will always be consistent. Consider the following system : ${\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}=0$ $a_{2}x+b_{2}y=0$ As the constant terms are zero, equation (1) and (2) will pass through the origin. As such, they will always intersect at one point i.e. the origin. Hence, such a system will always be consistent.

Building Concepts 5 Consider the following system of equations: $2x+5y=0$ and $4x+3y=0$ (a) If the system is consistent, how many solutions are possible, find them. (b) If the coefficient of $y$ in second equation is replaced by 10 , will there be any change in the number of solutions? Explain your answer. Explanation (a) $2x+5y=0$... (1) ; $4x+3y=0$... (2) Hence, $a_1=2,b_1=5,a_2=4,b_2=3$. $\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{2}{4}=\frac{1}{2}$ and $\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{5}{3}\Rightarrow \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}\ne \frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}$ $\Rightarrow$ Given system is consistent and has unique solution. $2x+5y=0\Rightarrow y=\frac{-2x}{5}4x+3y=0\Rightarrow y=\frac{-4x}{3}$

Points are $(5,-2),(-5,2),(0,0)$

From the graph, we see that the two lines intersect at a point $(0,0)$. So, the solution of the pair of linear equations are $x=0,y=0$. (b) On replacing the coefficient of y in second equation by 10 , we have :

$$\begin{gathered} 2x+5y=0 \\ 4x+10y=0 \end{gathered}$$ Here $\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{2}{4}=\frac{1}{2}$ and $\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{5}{10}=\frac{1}{2}$ $\Rightarrow \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}$ $\Rightarrow$ Now the system has infinite number of solutions.

$2x+5y=0$ $y=\frac{-2x}{5}$

Points are $(0,0),(5,-2),(-5,2)$

$4x+10y=0$ $y=\frac{-2x}{5}$

Points are $(0,0),(5,-2),(-5,2)$

From the graph (see fig.), we see that the two lines are coincident. So, the system has infinitely many solutions.

5.0Algebraic solution by substitution method

To solve a pair of linear equations in two variables x and y by substitution method, we follow the following steps:

Step-1: Write the given equations $$\begin{gathered} {\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0 \\ \text{ and }{\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2=0 \end{gathered}$$ Step- 2: Choose one of the two equations and express y in terms of x (or x in terms of y ), i.e., express one variable in terms of the other. Step-3: Substitute this value of $y$ obtained in step-2, in the other equation to get a linear equation in x . Step-4: Solve the linear equation obtained in step-3 and get the value of $x$. Step-5 : Substitute this value of x in the relation obtained in step- 2 and find the value of y .

Numerical Ability 4 Solve for x and $\mathrm{y}:4\mathrm{x}+3\mathrm{y}=24,3\mathrm{y}-2\mathrm{x}=6$. Solution: $4x+3y=24$ $3y-2x=6$ From equation (1), we get $y=\frac{24-4x}{3}$ Substituting in equation (2), we get $3\left(\frac{24-4x}{3}\right)-2x=6$ $\Rightarrow 24-4\mathrm{x}-2\mathrm{x}=6$ $\Rightarrow -6x=-24+6$ $\Rightarrow 6\mathrm{x}=18$ $\Rightarrow \mathrm{x}=3$ Substituting $x=3$ in (3), we get $\Rightarrow \mathrm{y}=\frac{24-12}{3}\Rightarrow \frac{12}{3}=4$ Hence, $x=3,y=4$.

A homogeneous system is always consistent.

Building Concepts 6 In one day the Museum admitted 321 adults and children and collected ₹ 1590 . The price of admission is ₹ 6 for an adult and ₹ 4 for a child. How many adults and how many children were admitted to the museum that day? Explanation:

• While applying substitution method, do not put the value of $y$ (in term of $x$ ) in the same equation from which the value of $y$ is obtained in terms of $x$.

6.0Algebraic solution by elimination method

To solve a pair of linear equations in two variables $x$ and $y$ by elimination method, we follow the following steps: Step-1: Write the given equations. $$\begin{gathered} {\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0 \\ \text{ and }{\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2=0 \end{gathered}$$ Step-2: Multiply the given equations by suitable numbers so that the coefficient of one of the variables are numerically equal. Step-3: If the numerically equal coefficients are opposite in sign, then add the new equations otherwise subtract. Step-4: Solve the linear equations in one variable obtained in step-3 and get the value of one variable. Step-5 : Substitute this value of the variable obtained in step-4 in any of the two equations and find the value of the other variable.

Numerical Ability 6 Solve the following pair of linear equations by elimination method $3x+4y=10$ and $2\mathrm{x}-2\mathrm{y}=2$. Solution We have, $3x+4y=10$ and $2x-2y=2$ Multiplying equation (2) by 2 , we get $4x-4y=4$ Adding (1) and (3), we get $7\mathrm{x}=14\Rightarrow \mathrm{x}=2$ Putting $x=2$ in equation (2), we get $2\times 2-2y=2$ $\Rightarrow 2\mathrm{y}=4-2$ $\Rightarrow y=1$ Hence, the solution is $\mathrm{x}=2$ and $\mathrm{y}=1$.

• In elimination method while multiplying any one (or both) the equations by the constant, multiplication is done with each term of the equations.

7.0Algebraic solution by cross-multiplication method

Consider the system of linear equations. ${\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0$ $a_{2}x+b_{2}y+c_2=0$ To solve it by cross multiplication method, we follow the following steps : Step-1 : Write the coefficients as follows :

$$\begin{gathered} \frac{x}{b_1{c}_2-b_2{c}_1}=\frac{y}{c_1{a}_2-c_2{a}_1}=\frac{1}{a_1{b}_2-a_2{b}_1} \\ \Rightarrow x=\frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1}\text{ and }y=\frac{c_1{a}_2-c_2{a}_1}{a_1{b}_2-a_2{b}_1} \end{gathered}$$

Case-1 : If $a_1{b}_2-a_2{b}_1\ne 0\Rightarrow x$ and $y$ have some finite values, with unique solution for the system of equations.

Case-2 : If ${\mathrm{a}}_1{\mathrm{b}}_2-{\mathrm{a}}_2{\mathrm{b}}_1=0\Rightarrow \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}$ Here two cases arise : (a) If $\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}=\lambda (\lambda \ne 0)$. Then ${\mathrm{a}}_1={\mathrm{a}}_2\lambda ,{\mathrm{b}}_1={\mathrm{b}}_2\lambda ,{\mathrm{c}}_1={\mathrm{c}}_2\lambda$

Put these values in equation ${\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0$ $\Rightarrow {\mathrm{a}}_2\lambda \mathrm{x}+{\mathrm{b}}_2\lambda \mathrm{y}+{\mathrm{c}}_2\lambda =0$ $\Rightarrow \lambda \left({\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2\right)=0$ but $\lambda \ne 0$ $\Rightarrow {\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2=0$ So, (i) and (ii) are dependent, so there are infinite number of solutions.

(b) If $\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}\ne \frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}$ $\Rightarrow \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\lambda (\lambda \ne 0)$. Then ${\mathrm{a}}_1={\mathrm{a}}_2\lambda ,{\mathrm{b}}_1={\mathrm{b}}_2\lambda$ Put these values in equation $a_{1}x+b_{1}y+c_1=0$ $\Rightarrow a_2\lambda x+b_2\lambda y+c_1=0$ $\Rightarrow \lambda \left({\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}\right)+{\mathrm{c}}_1=0$ but $\lambda \ne 0$ $\Rightarrow a_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}=-{\mathrm{c}}_2$ [from equation (ii)] Put this value in equation $\lambda \left(a_{2}x+b_{2}y\right)+c_1=0$, we get $\Rightarrow \lambda \left(-{\mathrm{c}}_2\right)+{\mathrm{c}}_1=0$ $\Rightarrow \lambda =\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}$ As $\lambda \ne \frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}$ and we are getting $\lambda =\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}$ So, system of equations is inconsistent and it has infinite many solution.

• In cross multiplication method try to apply basic method rather than applying formula directly because it might possible that you are remembering wrong formula.

Numerical Ability 6 Solve by cross-multiplication method: $\mathrm{x}+2\mathrm{y}+1=0$ and $2\mathrm{x}-3\mathrm{y}-12=0$ Solution: We have, $x+2y+1=0$ and $2x-3y-12=0$ By cross-multiplication method, we have $$\begin{gathered} \stackrel{x}{\overbrace{2}} \\ \therefore \frac{x}{2\times (-12)-(-3)\times 1}=\frac{y}{1\times 2-(-12)\times 1}=\frac{1}{1\times (-3)-2\times 2} \\ \Rightarrow \frac{\mathrm{x}}{-24+3}=\frac{\mathrm{y}}{2+12}=\frac{1}{-3-4} \\ \Rightarrow x=\frac{-21}{-7}=3\text{ and }y=\frac{14}{-7}=-2 \end{gathered}$$ Hence, the solution is $\mathrm{x}=3$ and $\mathrm{y}=-2$.

Building Concepts 7 Bus fare from Bangalore bus stand, if we buy 2 tickets to Malleswarm and 3 tickets to Yeshwanthpur, the total cost is ₹ 46; but if we buy $3$ tickets to Malleswarm and 5 tickets to Yeshwanthpur the total cost is ₹ 74. Find the fares from Bangalore to Malleswarm and to Yeshwanthpur. Explanation:

• Before applying cross multiplication method write both linear equations in two variables in standard form.

Interchanged Coefficients Method

Equations of the form $ax+by=c$ and $bx+ay=d$, where $a\ne b$. To solve the equations of the form: $ax+by=c$ and $bx+ay=d$ where $a\ne b$, we follow the following steps : Step-1: Add (1) and (2) and obtain $(a+b)x+(b+a)y=c+d$, i.e., $x+y=\frac{c+d}{a+b}$ Step-2 : Subtract (2) from (1) and obtain $(a-b)x-(a-b)y=c-d$, $\text{ i.e., }x-y=\frac{c-d}{a-b}$ Step-3: Solve (3) and (4) to get x and y .

Building Concepts 8 Solve for x and $\mathrm{y}:47\mathrm{x}+31\mathrm{y}=63,31\mathrm{x}+47\mathrm{y}=15$. Solution We have, $47x+31y=63$ and $31x+47y=15$ Adding (1) and (2), we get : $$\begin{gathered} 78x+78y=78 \\ \Rightarrow x+y=1 \end{gathered}$$ Subtracting (2) from (1), we get : $$\begin{gathered} 16x-16y=48 \\ \Rightarrow x-y=3 \end{gathered}$$ Now, adding (3) and (4), we get : $$\begin{gathered} 2\mathrm{x}=4 \\ \Rightarrow \mathrm{x}=2 \end{gathered}$$ Putting $x=2$ in (3), we get : $2+y=1$ $\Rightarrow y=-1$ Hence, the solution is $x=2$ and $y=-1$

8.0Equations reducible to linear equations in two variables

Equations which contain the variables, only in the denominators, are called reciprocal equations. These equations can be of the following types and can be solved by the under mentioned method:

Type-I: $\frac{\mathrm{a}}{\mathrm{u}}+\frac{\mathrm{b}}{\mathrm{v}}=\mathrm{c}$ and $\frac{{\mathrm{a}}^{\prime }}{\mathrm{u}}+\frac{{\mathrm{b}}^{\prime }}{\mathrm{v}}={\mathrm{c}}^{\prime }$ (where $\mathrm{a},\mathrm{b},\mathrm{c},{\mathrm{a}}^{\prime },{\mathrm{b}}^{\prime },{\mathrm{c}}^{\prime }$ are real number). Put $\frac{1}{u}=x$ and $\frac{1}{v}=y$ and find the value of $x$ and $y$ by any method described earlier. Then $\mathrm{u}=\frac{1}{\mathrm{x}}$ and $\mathrm{v}=\frac{1}{\mathrm{y}}$ Type-II : $\mathrm{au}+\mathrm{bv}=\mathrm{cuv}$ and a'u + b'v = c'uv (where $\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{a}$ ', ${\mathrm{b}}^{\prime },\mathrm{c}$ ' are real number). Divide both equations by uv and equations can be converted in the form explained in (I). Type-III : $\frac{a}{\mathrm{lx}+\mathrm{my}}+\frac{\mathrm{b}}{\mathrm{cx}+\mathrm{dy}}=\mathrm{k},\frac{{\mathrm{a}}^{\prime }}{\mathrm{lx}+\mathrm{my}}+\frac{{\mathrm{b}}^{\prime }}{\mathrm{cx}+\mathrm{dy}}={\mathrm{k}}^{\prime }$ (where $\mathrm{a},\mathrm{b},\mathrm{k},{\mathrm{a}}^{\prime },{\mathrm{b}}^{\prime },{\mathrm{k}}^{\prime }$ are real number) Put $\frac{1}{\mathrm{lx}+\mathrm{my}}=u$ and $\frac{1}{cx+dy}=v$ Then equations are $au+bv=k$ and $a^{\prime }u+b^{\prime }v=k^{\prime }$ Find the values of $u$ and $v$ and put in $lx+my=\frac{1}{u}$ and $cx+dy=\frac{1}{v}$ Again, solve for x and y , by any method explained earlier.

Numerical Ability 7 Solve for $x$ and $y:\frac{3a}{x}-\frac{2b}{y}+5=0$ and $\frac{a}{x}+\frac{3b}{y}-2=0(x\ne 0,y\ne 0)$ Solution We have, $\frac{3\mathrm{a}}{\mathrm{x}}-\frac{2\mathrm{b}}{\mathrm{y}}+5=0$ and $\frac{\mathrm{a}}{\mathrm{x}}+\frac{3\mathrm{b}}{\mathrm{y}}-2=0$ Let $\frac{1}{x}=u$ and $\frac{1}{y}=v$. Then, the given equations can be written as $3au-2bv=-5$ and $au+3bv=2$ Multiplying (1) by 3 and (2) by 2 , we get $9\mathrm{au}-6\mathrm{bv}=-15$ and $2\mathrm{au}+6\mathrm{bv}=4$ Adding (3) and (4), we get $11\mathrm{au}=-11$ $\Rightarrow \mathrm{u}=\frac{-1}{\mathrm{a}}$ $$\begin{gathered} \text{ Put }u=\frac{-1}{\mathrm{a}}\text{ in equation (2), we get }\mathrm{a}\left(\frac{-1}{\mathrm{a}}\right)+3\mathrm{bv}=2 \\ \Rightarrow 3\mathrm{bv}=3\Rightarrow \mathrm{v}=\frac{1}{\mathrm{b}}\text{ But }\frac{1}{\mathrm{x}}=\mathrm{u}\text{ and }\frac{1}{\mathrm{y}}=\mathrm{v} \end{gathered}$$ Therefore, $\frac{1}{\mathrm{x}}=\frac{-1}{\mathrm{a}}\Rightarrow \mathrm{x}=-\mathrm{a}$ and $\frac{1}{\mathrm{y}}=\frac{1}{\mathrm{b}}$ $\Rightarrow \mathrm{y}=\mathrm{b}\left[\because \mathrm{u}=\frac{-1}{\mathrm{a}},\mathrm{v}=\frac{1}{\mathrm{b}}\right]$ Hence the solution is $\mathrm{x}=-\mathrm{a}$ and $\mathrm{y}=\mathrm{b}$.

Numerical Ability 8 Solve for x and $\mathrm{y}:7\mathrm{x}-2\mathrm{y}=5\mathrm{xy}$ and $8\mathrm{x}+7\mathrm{y}=15\mathrm{xy}$ Solution: $7x-2y=5xy\Rightarrow \frac{7}{y}-\frac{2}{x}=5$ $8x+7y=15xy\Rightarrow \frac{8}{y}+\frac{7}{x}=15$ Putting $\frac{1}{\mathrm{y}}=\mathrm{u}$ and $\frac{1}{\mathrm{x}}=\mathrm{v}$, we get $7u-2v=5$ $8u+7v=15$ Multiplying (3) by 7 and (4) by 2 and adding, we get $49u-14v=35$ and $16u+14v=30$ $65u=65$ $\Rightarrow u=1$ $\Rightarrow \frac{1}{\mathrm{y}}=1$ or $\mathrm{y}=1$ Substituting $u=1$ in (3) we get : $7-2v=5\Rightarrow v=1\Rightarrow \frac{1}{x}=1$ or $x=1$ Hence, $x=1,y=1$.

Numerical Ability 9 Solve : $\frac{57}{x+y}+\frac{6}{x-y}=5$ and $\frac{38}{x+y}+\frac{21}{x-y}=9$. Solution: We have $\frac{57}{x+y}+\frac{6}{x-y}=5$ $\Rightarrow \frac{57}{x+y}+\frac{6}{x-y}-5=0$ and $\frac{38}{x+y}+\frac{21}{x-y}=9$ $\Rightarrow \frac{38}{x+y}+\frac{21}{x-y}-9=0$ Let $\frac{1}{x+y}=p$ and $\frac{1}{x-y}=q$ Then, the given equations can be rewritten as $57p+6q-5=0$ and $38p+21q-9=0$ By cross-multiplication method, we have $$\begin{gathered} \text{ ( }\stackrel{p}{\overbrace{6\times (-9)-21\times (-5)}}=\frac{p}{(-5)\times 38-(-9)\times 57}=\frac{q}{57\times 21-38\times 6} \\ \Rightarrow \frac{p}{-54+105}=\frac{q}{-190+513}=\frac{1}{1197-228} \\ \Rightarrow \frac{p}{51}=\frac{q}{323}=\frac{1}{969} \\ \Rightarrow p=\frac{51}{969}=\frac{1}{19}\text{ and }q=\frac{323}{969}=\frac{1}{3} \\ \text{ But }\frac{1}{x+y}=p\text{ and }\frac{1}{x-y}=q.\text{ Therefore }\frac{1}{x+y}=\frac{1}{19} \\ \Rightarrow x+y=19 \\ \text{ and }\frac{1}{x-y}=\frac{1}{3} \\ \Rightarrow x-y=3 \end{gathered}$$ Adding (3) and (4), we get $2x=22\Rightarrow x=11$ Put $x=11$ in (3), we get $11+y=19\Rightarrow y=8$ Hence, the solution is $\mathrm{x}=11$ and $\mathrm{y}=8$.

• Word problems are mainly dependent on formation of equations i.e. decoding of language given in word problem. Word Problems based on articles and their costs

Building Concepts 9 37 pens and 53 pencils together cost ₹ 320 , while 53 pens and 37 pencils together cost $₹400$. Find the cost of a pen and that of a pencil. Explanation: Let the cost of a pen be ₹ $x$ and that of a pencil be ₹ $y$. Then, $37x+53y=320$ and $53\mathrm{x}+37\mathrm{y}=400$ Adding equations (1) and (2), we get $90\mathrm{x}+90\mathrm{y}=720\Rightarrow \mathrm{x}+\mathrm{y}=8$ Subtracting equation (1) from (2), we get $16\mathrm{x}-16\mathrm{y}=80\Rightarrow \mathrm{x}-\mathrm{y}=5$ Adding equations (3) and (4), we get $2\mathrm{x}=13\Rightarrow \mathrm{x}=6.5$ Substituting $x=6.5$ in equation (3), we get $y=(8-6.5)=1.5$ Hence, cost of one pen = ₹ 6.50 and cost of one pencil $=₹1.50$.

Numerical Ability 10 $A$ and $B$ each have certain number of oranges. A says to $B$, "if you give me 10 of your oranges, I will have twice the number of oranges left with you". B replies, "if you give me 10 of your oranges, I will have the same number of oranges as left with you". Find the number of oranges with A and B separately. Solution Suppose A has x oranges and B has y oranges. According to the given conditions, we have $x+10=2(y-10)\Rightarrow x-2y+30=0$ and, $y+10=x-10\Rightarrow x-y-20=0$ Subtracting equation (2) from equation (1), we get $-y+50=0\Rightarrow y=50$ Putting $y=50$ in equation (1), we get $x=70$ Hence, A has 70 oranges and B has 50 oranges.

Word Problems based on numbers

• Recall that the two-digit number having a and b as units and ten's digits respectively is equal to $10\mathrm{b}+\mathrm{a}$ and the number obtained by reversing the order of digits is $10\mathrm{a}+\mathrm{b}$.

Building Concepts 10 In a two-digit number, the unit's digit is twice the ten's digit. If 27 is added to the number, the digits interchange their places. Find the number. Explanation: Let the digit in the unit's place be x and digit in the ten's place be y . Then, $\mathrm{x}=2\mathrm{y}$ [Given] and, Number $=10y+x$ Number obtained by reversing the digits $=10x+y$ It is given that the digits interchange their places if 27 is added to the number. i.e., Number $+27=$ Number obtained by interchanging the digits $\therefore 10\mathrm{y}+\mathrm{x}+27=10\mathrm{x}+\mathrm{y}$ $\Rightarrow 9x-9y=27$ $\Rightarrow \mathrm{x}-\mathrm{y}=3$ Putting $x=2y$ in equation (2), we get $2y-y=3\Rightarrow y=3$ Putting $y=3$ in equation (2), we get $\mathrm{x}=6$ Hence, the number is $10y+x=10\times 3+6=36$

Numerical Ability 11 The sum of a two digit number and the number obtained by reversing the order of its digits is 121 , and the two digits differ by 3 . Find the number. Solution: Let the digit in the units place be x and the digit at the ten's place be y . Then, number $=10\mathrm{y}+\mathrm{x}$ The number obtained by reversing the order of the digits is $10x+y$. According to the given conditions, we have $(10y+x)+(10x+y)=121$ $\Rightarrow 11(x+y)=121$ $\Rightarrow x+y=11$ and $\mathrm{x}-\mathrm{y}=\pm 3[\because$ Difference of digits is 3 ] Thus, we have the following sets of simultaneous equations. $$\begin{gathered} x+y=11 \\ x+y=11 \end{gathered}$$ and $x-y=3$ or $x-y=-3$ On solving equation (1) and (2), we get $x=7,y=4$ On solving equations (3) and (4), we get $x=4,y=7$ When $\mathrm{x}=7,\mathrm{y}=4$, we have Number $=10y+x=10\times 4+7=47$ When $x=4,y=7$, we have Number $=10y+x=10\times 7+4=74$ Hence, the required number is either 47 or 74 . Word Problems based on fractions

Building Concepts 11 A fraction is such that if the numerator is multiplied by 3 and the denominator is reduced by 3 , we get $\frac{18}{11}$, but if the numerator is increased by 8 and the denominator is doubled, we get $\frac{2}{5}$. Find the fraction. Explanation: Let the fraction be $\frac{x}{y}$. Then, according to the given conditions, we have $$\begin{gathered} \frac{3x}{y-3}=\frac{18}{11}\text{ and }\frac{x+8}{2y}=\frac{2}{5} \\ \Rightarrow 11x=6y-18\text{ and }5x+40=4y \\ \Rightarrow 11x-6y+18=0 \\ 5x-4y+40=0 \end{gathered}$$ Multiplying equation (1) by 2 and (2) by 3 and subtracting (2) from (1), $22\mathrm{x}-12\mathrm{y}+36=0$ $15x-12y+120=0$ $7\mathrm{x}-84=0$ $x=\frac{84}{7}=12$ Putting $x=12$ in equation (2) $5\times 12-4y+40=0$ $60+40=4y$ $y=\frac{100}{4}=25$ Hence, the fraction is $\frac{12}{25}$.

Numerical Ability 12 The denominator of a fraction is 4 more than twice the numerator. When both the numerator and denominator are decreased by 6, then the denominator becomes 12 times the numerator. Determine the fraction. Solution: Let the numerator and denominator of the fraction be x and y respectively. Then, Fraction $=\frac{x}{y}$ It is given that, Denominator $=2$ (Numerator) +4 $\Rightarrow y=2x+4$ $\Rightarrow 2\mathrm{x}-\mathrm{y}+4=0$ According to the given condition, we have $$\begin{gathered} \mathrm{y}-6=12(\mathrm{x}-6) \\ \Rightarrow \mathrm{y}-6=12\mathrm{x}-72 \\ \Rightarrow 12\mathrm{x}-\mathrm{y}-66=0 \end{gathered}$$ Thus, we have the following system of equations. $$\begin{gathered} 2x-y+4=0 \\ 12x-y-66=0 \end{gathered}$$ Subtracting equation (1) from equation (2), we get $10x-70=0\Rightarrow x=7$ Putting $x=7$ in equation (1), we get $14-y+4=0\Rightarrow y=18$ Hence, required fraction $=\frac{7}{18}$.

Word Problems based on ages

Building Concepts 12 If twice the son's age in years is added to the father's age, the sum is 70. But if twice the father's age is added to the son's age, the sum is 95 . Find the ages of father and son. Explanation: Suppose father's age (in years) be $x$ and that of son's age be $y$. Then, $x+2y=70$ and $2x+y=95$ This system of equations may be written as $\mathrm{x}+2\mathrm{y}-70=0$ $2\mathrm{x}+\mathrm{y}-95=0$ Multiplying equation (1) by 2 and subtracting from (2) $2\mathrm{x}+\mathrm{y}-95=0$ $2\mathrm{x}+4\mathrm{y}-140=0$ $-3y+45=0$ $\mathrm{y}=15$ Putting in equation (1) $x+2\times 15-70=0$ $\mathrm{x}=40$ Hence, father's age is 40 years and the son's age is 15 years.

Numerical Ability 13 Ten years ago, father was twelve times as old as his son and ten years hence, he will be twice as old as his son. Find their present ages. Solution: Let present age of father and his son (in years) be $x$ and $y$ respectively. Ten years ago, Father's age $=(x-10)$ years Son's age $=(y-10)$ years $\therefore x-10=12(y-10)$ $\Rightarrow x-12y+110=0$ Ten years later, Father's age $=(x+10)$ years. Son's age $=(y+10)$ $\therefore x+10=2(y+10)$ $\Rightarrow x-2y-10=0$ Subtracting (2) from (1), we get $-10y+120=0$ $\Rightarrow 10\mathrm{y}=120$ $\Rightarrow y=12$ Putting $y=12$ in (1), we get $x-144+110=0$ $\Rightarrow \mathrm{x}=34$ Thus, present age of father is 34 years and the present age of son is 12 years.