Question

Solve the following equations by the substitution method :

$$\begin{gathered} \left(vi\right)8x-5y+40=0 \\ 7x-2y=0 \end{gathered}$$

Answer 1

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

$$\begin{gathered} 7x-2y=0 \\ 7x=2y \\ x=\frac{2}{7}y \end{gathered}$$

Step 2: Substitute this value of x in the other equations to get a linear equation in y which can be solved:

$$\begin{gathered} 8x-5y+40=0 \\ 8\left(\frac{2}{7}y\right)-5y+40=0 \end{gathered}$$

Multiplying the linear equation throughout by 7:

$$\begin{gathered} 8\times 2y-35y+280=0 \\ 16y-35y=-280 \\ -19y=-280 \\ y=\frac{280}{19} \end{gathered}$$

Step 3: Substitute the values of y obtained in $x=\frac{2y}{7}$ to get the values of x:

$$\begin{gathered} x=\frac{2}{7}y \\ =\frac{2}{7}\times \frac{280}{19} \\ =\frac{2\times 40}{19} \\ =\frac{80}{19} \end{gathered}$$

Hence, the value of $x=\frac{80}{19}andy=\frac{280}{19}$.