Practice Problems

Answers

  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:

    \(x=9\)

    \(y=12\)