Question

Show and explain how replacing one equation by thesum of that equation and a multiple of the other producesa system with the same solutions as the one shown.8x + 7y = 394x – 14y = -68

Answer 1

Given a system of equation:

`\begin{bmatrix}8x+7y=39 \\ 4x-14y=-68\end{bmatrix}`

Step 1: Isolate the equation 1 for x

`\begin{gathered} 8x+7y=39 \\ 8x=39-7y \\ x=(39-7y)/(8) \end{gathered}`

Step 2: Substitute the value of x in equation 2

`\begin{gathered} 4x-14y=-68 \\ 4((39-7y)/(8))-14y=-68 \\ (39-7y)/(2)-14y=-68 \\ \text{ multiply through by 2} \\ 39-7y-28y=-126 \\ 39-35y=-126 \\ -35y=-136-39 \\ -35y=-175 \\ y=-(175)/(-35) \\ y=5 \end{gathered}`

Step 3: Substitute the value of y

`\begin{gathered} \mathrm{Substitute\: }y=5 \\ x=(39-7y)/(8) \\ x=(39-7(5))/(8) \\ x=(39-35)/(8) \\ x=(4)/(8) \\ x=(1)/(2) \end{gathered}`

Hence the correct answer for the system of equations are

`x=(1)/(2),y=5`