Question

Solve the following system of equations by the elimination method and the substitution method:

$$\begin{gathered} 2y-x=6 \\ 3y-2x=10 \end{gathered}$$

Answer 1

Elimination method

Step(1): Making coefficient of variable$(xory)$equal

Multiply the given equation needed by a suitable number so that the coefficient of one of the variables says$(xory)$ become equal.

Multiply by $2$ in $2y-x=6$ which make $x$coefficient equal.

$$\begin{gathered} 4y-2x=12 \\ 3y-2x=10 \end{gathered}$$

Step (2): Eliminating the variable $x$

Subtract the equation obtained in step$-\left(1\right)$ to eliminate one variable. The resulting equation is a linear equation in one variable.

$4y-3y=12-10$

$\Rightarrow y=2$

Step (3): Obtain value of other variable

Substitute this value of the variable in either of the original equations, to get the value of the other variable.

$$\begin{gathered} 4y-2x=12 \\ \Rightarrow (4\times 2)-2x=12 \\ \Rightarrow 8-2x=12 \\ \Rightarrow -2x=12-8 \\ \Rightarrow x=\frac{4}{-2} \\ \Rightarrow x=-2 \end{gathered}$$

Hence $x=-2\&y=2$

Substitution Method

Step(1):Express one variable in terms of other

Express one variable $(sayx)$ in terms of other variable $(sayy)$ from one of the given equations.

$$\begin{gathered} 2y-x=6 \\ 2y-6=x \end{gathered}$$

Step (2):Use the second equation

Substitute this value of $x$ in the other equations to get a linear equations in $y$ which can be solved

$$\begin{gathered} 3y-2x=10 \\ \Rightarrow 3y-2(2y-6)=10 \\ \Rightarrow 3y-4y+12=10 \\ \Rightarrow -y=10-12 \\ \Rightarrow y=2 \end{gathered}$$

Step (3) : Finding the other variable.

Substitute the values of $y$ obtained in step-$\left(2\right)$ above in the equations used in step- $\left(1\right)$to get the values of $x$.

$$\begin{gathered} x=2y-6 \\ \Rightarrow x=(2\times 2)-6 \\ \Rightarrow x=4-6 \\ \Rightarrow x=-2 \end{gathered}$$

Hence,by both mehod answer is same and $x=-2\&y=2$