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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

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Answer:

The solution to the system of equation is;


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

Step-by-step 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=(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=(30)/(2) \\ y=15 \end{gathered}

Therefore, the solution to the system of equation is;


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

User Douglas Ribeiro
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