Question

Solve the pairs of linear equation by the elimination method and substitution method:
 \[3x - 5y - 4 = 0\] and \[9x = 2y + 7\]

Answer 1

Hint: Given, the linear equation in two variables. We need to solve the linear equation in both elimination method and substitution method. In elimination method we make the coefficient of one of the variables in two linear equations the same by multiplying suitable numbers. In the substitution method we change the linear equation in one variable by using one of the equations.

Complete step-by-step answer:
I.Rearranging the given equation, we have
 \[3x - 5y = 4\] ---- (1) \[9x - 2y = 7\] ----- (2)
Now we do elimination method:
As we know we need to make the coefficient of one of the variables same.
Now multiply whole equation by 3
 \[\left( {3x - 5y = 4} \right) \times 3\]
 \[ \Rightarrow 9x - 15y = 12\] ----- (3)
Subtract equation (2) with equation (3), we have,
 \[ \Rightarrow 9x - 2y - (9x - 15y) = 7 - 12\]
 \[ \Rightarrow 9x - 2y - 9x + 15y = - 5\]
 \[ \Rightarrow - 2y + 15y = - 5\]
 \[ \Rightarrow 13y = - 5\]
 \[ \Rightarrow y = - \dfrac{5}{{13}}\]
To find the value of x, substitute in y value in equation (2)
 \[ \Rightarrow 9x - 2\left( { - \dfrac{5}{{13}}} \right) = 7\]
 \[ \Rightarrow 9x + \dfrac{{2 \times 5}}{{13}} = 7\]
 \[ \Rightarrow 9x + \dfrac{{10}}{{13}} = 7\]
Rearranging we have,
 \[ \Rightarrow 9x = 7 - \dfrac{{10}}{{13}}\]
 \[ \Rightarrow 9x = \dfrac{{13 \times 7 - 10}}{{13}}\]
 \[ \Rightarrow 9x = \dfrac{{91 - 10}}{{13}}\]
 \[ \Rightarrow 9x = \dfrac{{81}}{{13}}\]
Divide by 9 on both sides,
 \[ \Rightarrow x = \dfrac{9}{{13}}\] .
Thus by elimination method we have \[x = \dfrac{9}{{13}}\] and \[y = - \dfrac{5}{{13}}\]

II.Now by substitution method,
Rearranging equation (1) we have,
 \[ \Rightarrow 3x = 4 + 5y\]
 \[ \Rightarrow x = \dfrac{{4 + 5y}}{3}\]
Now substituting x in the equation (2), we have
 \[9x - 2y = 7\]
 \[ \Rightarrow 9 \times \left( {\dfrac{{4 + 5y}}{3}} \right) - 2y = 7\]
 \[ \Rightarrow 3 \times (4 + 5y) - 2y = 7\]
 \[ \Rightarrow 12 + 15y - 2y = 7\]
 \[ \Rightarrow 12 + 13y = 7\]
 \[ \Rightarrow 13y = 7 - 12\]
 \[ \Rightarrow 13y = - 5\]
 \[ \Rightarrow y = - \dfrac{5}{{13}}\]
To get the value of x, substituting in equation (1) we have,
 \[ \Rightarrow 3x - 5 \times \left( { - \dfrac{5}{{13}}} \right) = 4\]
 \[ \Rightarrow 3x + \dfrac{{25}}{{13}} = 4\]
 \[ \Rightarrow 3x = 4 - \dfrac{{25}}{{13}}\]
 \[ \Rightarrow 3x = \dfrac{{13 \times 4 - 25}}{{13}}\]
 \[ \Rightarrow 3x = \dfrac{{52 - 25}}{{13}}\]
 \[ \Rightarrow 3x = \dfrac{{27}}{{13}}\]
Divide by 3 on both side, we have
 \[ \Rightarrow x = \dfrac{9}{{13}}\]
Thus by substitution method we have \[x = \dfrac{9}{{13}}\] and \[y = - \dfrac{5}{{13}}\]
So, the correct answer is “ \[x = \dfrac{9}{{13}}\] and \[y = - \dfrac{5}{{13}}\] ”.

Note: Thus we can see that in the both methods we have the same answer. Follow the same procedure for any given pair of linear equations with two variables. We can also do this by the method of cross multiplication. Know the difference between substitution and elimination method. In the substitution method we can also substitute the value of x instead of x (see in above). Either way we will get the same answer.