
Practice Problems
Answers
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We are given that \(y = {2x-5}\). Let's substitute this expression into the first equation and solve for \(x\) as follows:
\(\begin{aligned} 5x-4{y}&=-10\\\\
5x-4\cdot({2x-5})&=-10\\\\
5x-8x+20& = -10\\\\
-3x&=-30\\\\
x&=10
\end{aligned}\)Since we now know that \(x=10\), we can substitute this value in the second equation to solve for \(y\) as follows:
\(\begin{aligned} y &= 2\cdot {x}-5 \\\\
y&=2\cdot{10}-5\\\\
y&=15 \end{aligned}\)This is the solution of the system:
\(x=10\)
\(y=15\)
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We are given that \({x}={2y}\). Let's substitute this expression into the first equation and solve for \(y\) as follows:
\( \begin{aligned} -4{x}+11y &= 15\\\\
-4\cdot{2y}+11y&=15\\\\
-8y+11y&=15\\\\
3y&=15\\\\
y&=5
\end{aligned}\)Since we now know that \({y}={5}\), we can substitute this value in the second equation to solve for \(x\) as follows:
\(\begin{aligned} x &= 2\cdot{y} \\\\
x&=2\cdot{5}\\\\
x&=10 \end{aligned}\)This is the solution of the system:
\(x=10\)
\(y=5\)
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We are given that \(y = {x-2}\). Let's substitute this expression into the first equation and solve for \(x\) as follows:
\( \begin{aligned} 10x-9{y}&=24\\\\
10x-9\cdot({x-2})&=24\\\\
10x-9x+18& = 24\\\\
x&=6
\end{aligned}\)Since we now know that \({x}={6}\), we can substitute this value in the second equation to solve for \(y\) as follows:
\(\begin{aligned} y &= {x}-2 \\\\
y&={6}-2\\\\
y&=4 \end{aligned}\)This is the solution of the system:
\(x=6\)
\(y=4\)
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We are given that \({x}={2y-15}\). Let's substitute this expression into the first equation and solve for \(y\) as follows:
\(\begin{aligned} -5{x}+4y &= 3\\\\
-5\cdot({2y-15})+4y&=3\\\\
-10y+75+4y&=3\\\\
-6y&=-72\\\\
y&=12
\end{aligned}\)Since we now know that \({y}={12}\), we can substitute this value in the second equation to solve for \(x\) as follows:
\(\begin{aligned} x &= 2\cdot{y}-15 \\\\
x&=2\cdot{12}-15\\\\
x&=9 \end{aligned}\)This is the solution of the system:
\(x=9\)
\(y=12\)