Question

Solve by using method of elimination by substitution​

Answer 1

$\LARGE{\bf{\underline{\underline{Solution:-}}}}$

Step 1 : From any of the given two equations, we will find the value of one variable in terms of the other.

Let us find the value of x:

$\dfrac{2x+1}{7} +\dfrac{5y-3}{3}=12$

$\to\dfrac{6x+3+35y-21}{21} =12$

$\to\dfrac{6x+35y-18}{21} =12$

$\to \sf{{6x+35y-18=252}}$

$\to \sf{{6x+35y=270}}$

$\to \sf{{6x=270-35y}}$

$\to\sf{{x=}}$ $\dfrac{270-35y}{6}$

Step 2 : Substitute the value of variable, obtained in step ( 1 ), in the other equation and solve it.

$\dfrac{3x+2}{2} -\dfrac{4y+3}{9}=13$

$\to\dfrac{27x+18-(8y+6)}{18}=13$

$\to\sf{{27x-8y+12=234}}$

$\to\sf{{27x-8y=222}}$

Let us put the value of x which we obtained in step ( 1 )

$27(\dfrac{270-35y}{6})-8y=222$

$\to\dfrac{7290-945y}{6} -\dfrac{8y}{1}=222$

$\to\dfrac{7290-945y-48y}{6}=222$

$\to \sf{{7290-993y=222×6}}$

$\to\sf{{7290-993y=1332}}$

$\to\sf{{-993y=1332-7290}}$

$\to\sf{{-993y=-5850}}$

$\to\sf{{y}}=\cancel\dfrac{-5850}{-993}$

$\to\sf{{y=6}}$

Step 3 : Substitute the value of the variable obtained in step ( 2 ), in the result of step ( 1 ) and get the value of the remaining unknown variable.

$x=\dfrac{270-35y}{6}$

$\to\sf{{x}}=\dfrac{270-35(6)}{6}$

$\to\sf{{x}}=\dfrac{270-210}{6}$

$\to\sf{{x}}=\cancel\dfrac{60}{6}$

$\to\sf{{x=10}}$

$\LARGE{\bf{\underline{\underline{Verification:-}}}}$

$\dfrac{2x+1}{7} +\dfrac{5y-3}{3}=12$

$\to\dfrac{2(10)+1}{7} +\dfrac{5(6)-3}{3}=12$

$\to\dfrac{20+1}{7} +\dfrac{30-3}{3}=12$

$\to\dfrac{21}{7} +\dfrac{27}{3}=12$

$\to\sf{{3+9=12}}$

$\to\sf{{12=12}}$

$\to \sf{{L.H.S=R.H.S}}$

Thus, Value of x = 10

Value of y =6