Question

what is the solution to the system of equations ? 3r+2y=45 4r-y=5 1. r=5 y=10 2.r=15 y=10 3.r=5 y=15 4. r=15 y=5

Answer 1

Answer:

The solution to the system of equation is;

$\begin{gathered} r=5 \\ y=15 \end{gathered}$

Explanation:

Given the system of equations;

$\begin{gathered} 3r+2y=45\text{ -----------1} \\ 4r-y=5\text{ ----------2} \end{gathered}$

Let us solve by elimination, multiply equation 2 by 2 and add to equation 1 to eliminate y;

$\begin{gathered} 4r(2)-y(2)=5(2) \\ 8r-2y=10\text{ -----------3} \end{gathered}$

adding to equation 1;

$\begin{gathered} 3r+8r+2y-2y=45+10 \\ 11r=55 \\ r=\frac{55}{11} \\ r=5 \end{gathered}$

we can now use the value of r to solve for y;

$\begin{gathered} 3r+2y=45 \\ 3(5)+2y=45 \\ 15+2y=45 \\ 2y=45-15 \\ 2y=30 \\ y=\frac{30}{2} \\ y=15 \end{gathered}$

Therefore, the solution to the system of equation is;

$\begin{gathered} r=5 \\ y=15 \end{gathered}$