Question

I need help solving this please! Do I use substitution? Not sure how to solve?

Answer 1

Answer

$x=5,y=1$

Explanation

The system of the equation:

$\begin{gathered} 2x+3y=13\text{ ------(i)} \\ 4x-3y=17\text{ ------(}ii) \end{gathered}$

From (i)

$\begin{gathered} 2x=13-3y \\ \text{Divide both sides by 2} \\ \frac{2x}{2}=\frac{13-3y}{2} \\ x=\frac{13-3y}{2}\text{ -------(}iii) \end{gathered}$

Substitute x as (13 -3y)/2 into (ii)

$\begin{gathered} \text{Recall (}ii) \\ 4x-3y=17 \\ 4(\frac{13-3y}{2})-3y=17 \\ 2(13-3y)-3y=17 \\ 26-6y-3y=17 \\ 26-9y=17 \\ \text{Grouping the terms, we have} \\ 9y=26-17 \\ 9y=9 \\ y=\frac{9}{9} \\ y=1 \end{gathered}$

Since y = 1, then we shall substitute for y as 1 in (iii) to get x

$\begin{gathered} \text{Recall (}iii) \\ x=\frac{13-3y}{2} \\ x=\frac{13-3(1)}{2} \\ x=\frac{13-3}{2} \\ x=\frac{10}{2} \\ x=5 \end{gathered}$

Therefore the solution to the system is:

$x=5,y=1$