Question

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

Answer 1

`x=5,y=1`

Step-by-step 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} \\ (2x)/(2)=(13-3y)/(2) \\ x=(13-3y)/(2)\text{ -------(}iii) \end{gathered}`

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

`\begin{gathered} \text{Recall (}ii) \\ 4x-3y=17 \\ 4((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=(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=(13-3y)/(2) \\ x=(13-3(1))/(2) \\ x=(13-3)/(2) \\ x=(10)/(2) \\ x=5 \end{gathered}`

Therefore the solution to the system is:

`x=5,y=1`