Practice Using the Substitution Method

Solve the system of equations.

1. \(5x−4y=-10\)

   \(y=2x-5\)

2. \(-4x+11y=15\)

   \(x=2y\)

3. \(10x-9y=24\)

   \(y=x-2\)

4. \(-5x+4y=3\)

   \(x=2y-15\)

Source: Khan Academy, https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:systems-of-equations/x2f8bb11595b61c86:solving-systems-of-equations-with-substitution/e/systems_of_equations_with_substitution
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1. 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:

   \(x=10\)

   \(y=15\)
   

2. 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:

   \(x=10\)

   \(y=5\)

3. 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:

   \(x=6\)

   \(y=4\)

4. 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:

   \(x=9\)

   \(y=12\)