Question

Solve $2x-y=12$ and $x+3y+1=0$ and hence find the value of m for which $y=mx+3$.

Answer 1

Step(1):Express one variable (say x) in terms of other variable (say y) from one of the given equations.

Given equations,

$$\begin{gathered} 2x-y=12............................................\left(1\right) \\ x+3y+1=0.........................................\left(2\right) \end{gathered}$$

Consider the first equation,

$$\begin{gathered} 2x-y=12 \\ 2x=12+y \\ x=\frac{12+y}{2}.......................................\left(3\right) \end{gathered}$$

Step (2):substitute this value of $x=\frac{12+y}{2}$in equation (2) to get a linear equations in y which can be solved.

Consider the second equation,

$x+3y+1=0$

$\frac{12+y}{2}+3y+1=0$

multiplying the equation throughout by 2

$$\begin{gathered} 12+y+6y+2=0 \\ \Rightarrow 14+7y=0 \\ \Rightarrow 7y=-14 \\ \Rightarrow y=\frac{-14}{7} \\ \Rightarrow y=-2 \end{gathered}$$

Step (3) : Substitute the values of $y=-2$ in equation (3), we will get the values for x

$$\begin{gathered} x=\frac{12+y}{2} \\ \Rightarrow x=\frac{12-2}{2} \\ \Rightarrow x=\frac{10}{2} \\ \Rightarrow x=5 \end{gathered}$$

Step (4) : Substitute the values of x and y obtained in the expression $y=mx+3$ to get the values of m.

$$\begin{gathered} y=mx+3 \\ \Rightarrow -2=m\times 5+3 \\ \Rightarrow -2-3=5m \\ \Rightarrow -5=5m \\ \Rightarrow m=\frac{-5}{5} \\ \Rightarrow m=-1 \end{gathered}$$

Hence, the value of $mis-1.$