Question

By equating coefficients of variables, solve the following equations.

(i) 3x $-$ 4 y = 7; 5x + 2y = 3

(ii) 5x + 7 y = 17 ; 3x $-$2y = 4

(iii) x $-$2y = $-$10; 3x $-$ 5y = $-$12

(iv) 4x + y = 34 ; x + 4y = 16

Answer 1

(i) 3x $-$ 4 y = 7 .....(I)
5x + 2y = 3 .....(II)
Multiply (II) with 2
10x + 4y = 6 .....(III)
Adding (I) and (III)
$$\begin{gathered} 13x=13 \\ \Rightarrow x=1 \end{gathered}$$
Putting the value of x in (I) we get
$$\begin{gathered} 3\times 1-4y=7 \\ \Rightarrow 3-4y=7 \\ \Rightarrow y=-1 \end{gathered}$$
Thus, x = 1 and $y=-1$.

(ii) 5x + 7 y = 17 .....(I)
3x $-$2y = 4 .....(II)
Multiply (I) with 3 and (II) with 5
15x + 21 y = 51 .....(III)
15x $-$10y = 20 .....(IV)
Subtracting (IV) from (III) we get
$$\begin{gathered} 31y=31 \\ \Rightarrow y=1 \end{gathered}$$
Putting this value of y in (I) we get
$$\begin{gathered} 5x+7\times 1=17 \\ \Rightarrow 5x=10 \\ \Rightarrow x=2 \end{gathered}$$
Thus, $x=2,y=1$.

(iii) x $-$2y = $-$10 .....(I)
3x $-$ 5y = $-$12 .....(II)
Multiply (I) with 3
$3x-6y=-30$ .....(III)
Subtracting (II) from (III) we get
$$\begin{gathered} -y=-18 \\ \Rightarrow y=18 \end{gathered}$$
Putting the value of y in (I) we get
$$\begin{gathered} x-2\times 18=-10 \\ \Rightarrow x=-10+36=26 \end{gathered}$$
Thus, $x=26,y=18$

(iv) 4x + y = 34 .....(I)
x + 4y = 16 .....(II)
Adding (I) and (II) we get
$$\begin{gathered} 5x+5y=50 \\ \Rightarrow x+y=10.....\left(\mathrm{I}\right) \end{gathered}$$
Subtracting (II) from (I) we get
$$\begin{gathered} 3x-3y=18 \\ \Rightarrow x-y=6.....\left(\mathrm{IV}\right) \end{gathered}$$
(III) + (IV)
$$\begin{gathered} 2x=16 \\ \Rightarrow x=8 \end{gathered}$$
Putting the value of x in (I) we get
$$\begin{gathered} 4\left(8\right)+y=34 \\ \Rightarrow y=34-32=2 \\ \Rightarrow y=2 \end{gathered}$$
Thus, $x=8,y=2$