Question

5) Solve the following systems of linear equations: (10pts each) Sy=44-7 (use substitution) Ly=-2x+5 i) ii) (2x - 3y=8 (use elimination) 16x+5y = -4

Answer 1

The given set of equations are:

$\begin{gathered} 2x-3y=8\ldots\text{.}\mathrm{}(1) \\ 6x+5y=-4\ldots\text{.}\mathrm{}(2) \end{gathered}$

Multiply equation 1 by 5 and equation 2 by 3, we get,

$\begin{gathered} 5(2x-3y)=8\times5\rightarrow10x-15y=40\ldots\text{.}(3) \\ 3(6x+5y)=3\times(-4)\rightarrow18x+15y=-12\ldots\text{.}(4) \end{gathered}$

Now, add equation 3 and 4, we get,

$\begin{gathered} 10x-15y+18x+15y=40-12 \\ 28x=28 \\ x=1 \end{gathered}$

Therefore, from equation 1, with the value of x, we get,

$\begin{gathered} 2-3y=8 \\ 2-8=3y \\ -6=3y \\ y=-\frac{6}{3}=-2 \end{gathered}$

Thus, the solution is, x = 1 and y = -2