Question

Solve the system of equations below using substitution4x – 6y = 303x + 7y = 80

Answer 1

Solution:

The system of equation is given below as

$\begin{gathered} 4x-6y=30-------(1) \\ 3x+7y=80-------(2) \end{gathered}$

Step 1:

From equation (1), make x the subject of the formula

$\begin{gathered} 4x-6y=30 \\ 4x=30+6y \\ \frac{4x}{4}=\frac{30}{4}+\frac{6y}{4} \\ x=\frac{30}{4}+\frac{6y}{4}-------(3) \end{gathered}$

Step 2:

Substitute equation (3) in equation (2)

$\begin{gathered} 3x+7y=80 \\ 3(\frac{30}{4}+\frac{6y}{4})+7y=80 \\ \frac{90}{4}+\frac{18y}{4}+7y=80 \\ multiply\text{ through by 4} \\ 4(\frac{90}{4})+4(\frac{18y}{4})+4(7y)=4(80) \\ 90+18y+28y=320 \\ 46y+90=320 \\ 46y=320-90 \\ 46y=230 \\ \frac{46y}{46}=\frac{230}{46} \\ y=5 \end{gathered}$

Step 3:

Substitute y=5 in equation (3)

$\begin{gathered} x=\frac{30}{4}+\frac{6y}{4} \\ x=\frac{30}{4}+6(\frac{5}{4}) \\ x=\frac{30}{4}+\frac{30}{4} \\ x=\frac{60}{4} \\ x=15 \end{gathered}$

Hence,

The solution to the system of equations is

$\Rightarrow x=15,y=5$