Question

solve the system of equations.
\

$$\begin{cases} 5x - 5y = 20 \\\\ 3x - y = 11 \\end{cases}$$

both equations are in the same form, so you should use the linear combinations method.

1. write a new equation in a single variable.

determine whether the equations are ready to be combined.
one or both equations need to be modified before they can be combined using addition.
select the variable to eliminate: y

multiply the equation $5x - 5y = 20$ by \
ule{1cm}{0.15mm}
$\downarrow$
\
ule{1cm}{0.15mm} = \
ule{1cm}{0.15mm}

multiply the equation $3x - y = 11$ by \
ule{1cm}{0.15mm}
$\downarrow$
\
ule{1cm}{0.15mm} = \
ule{1cm}{0.15mm}

1. solve for one variable.

1. solve for the other variable.

Explanation:

Step1: Modify the second equation

To eliminate \( y \), we need the coefficients of \( y \) to be opposites. The first equation has \( -5y \), so we multiply the second equation \( 3x - y = 11 \) by \( 5 \) to get \( 15x - 5y = 55 \).
Multiply \( 3x - y = 11 \) by \( 5 \): \( 5\times(3x - y)=5\times11 \) → \( 15x - 5y = 55 \)

Step2: Add the two equations

Now we have the first equation \( 5x - 5y = 20 \) and the modified second equation \( 15x - 5y = 55 \). Wait, no, actually, if we want to eliminate \( y \), we should have one with \( -5y \) and one with \( +5y \). Wait, I made a mistake. Let's correct that. The first equation is \( 5x - 5y = 20 \), the second is \( 3x - y = 11 \). Let's multiply the second equation by \( -5 \) to get \( -15x + 5y = -55 \). Now add the first equation \( 5x - 5y = 20 \) and \( -15x + 5y = -55 \).

Adding them: \( (5x - 5y)+(-15x + 5y)=20+(-55) \) → \( 5x - 15x - 5y + 5y = -35 \) → \( -10x = -35 \)

Step3: Solve for \( x \)

Divide both sides by \( -10 \): \( x=\frac{-35}{-10}=\frac{7}{2}=3.5 \)

Step4: Solve for \( y \)

Substitute \( x = \frac{7}{2} \) into the second equation \( 3x - y = 11 \). \( 3\times\frac{7}{2}-y = 11 \) → \( \frac{21}{2}-y = 11 \) → \( -y = 11-\frac{21}{2}=\frac{22 - 21}{2}=\frac{1}{2} \) → \( y = -\frac{1}{2}=-0.5 \)

Wait, let's redo the elimination correctly. Let's choose to eliminate \( y \). The first equation: \( 5x - 5y = 20 \), second: \( 3x - y = 11 \). Let's multiply the second equation by \( 5 \): \( 15x - 5y = 55 \). Now subtract the first equation from this? No, the problem says using addition. Wait, the first equation has \( -5y \), the second (after multiplying by 5) has \( -5y \), so that's not helpful. So instead, multiply the second equation by \( -5 \): \( -15x + 5y = -55 \). Now add to the first equation \( 5x - 5y = 20 \):

\( (5x - 5y)+(-15x + 5y)=20 + (-55) \)

\( 5x - 15x - 5y + 5y = -35 \)

\( -10x = -35 \)

\( x=\frac{35}{10}=\frac{7}{2}=3.5 \)

Then substitute \( x = \frac{7}{2} \) into \( 3x - y = 11 \):

\( 3\times\frac{7}{2}-y = 11 \)

\( \frac{21}{2}-y = 11 \)

\( -y = 11 - \frac{21}{2}=\frac{22 - 21}{2}=\frac{1}{2} \)

\( y = -\frac{1}{2} \)

Wait, maybe the initial step was to multiply the second equation by 5, but then we have to adjust. Let's follow the problem's instruction: "Select the variable to eliminate: y. Multiply the equation \( 5x - 5y = 20 \) by [ ]? No, wait, the two equations are \( 5x - 5y = 20 \) and \( 3x - y = 11 \). To eliminate \( y \), we can multiply the second equation by 5 to get \( 15x - 5y = 55 \), then subtract the first equation from the second? But the problem says "using addition". Wait, maybe the problem has a typo, or I misread. Wait, the first equation: \( 5x - 5y = 20 \), second: \( 3x - y = 11 \). Let's write them:

Equation 1: \( 5x - 5y = 20 \)

Equation 2: \( 3x - y = 11 \)

Multiply Equation 2 by 5: \( 15x - 5y = 55 \) (Equation 3)

Now, if we add Equation 1 and Equation 3: \( (5x - 5y)+(15x - 5y)=20 + 55 \) → \( 20x - 10y = 75 \), which doesn't eliminate \( y \). So that's wrong. So we should multiply Equation 2 by -5: \( -15x + 5y = -55 \) (Equation 4)

Now add Equation 1 and Equation 4: \( (5x - 5y)+(-15x + 5y)=20 + (-55) \) → \( -10x = -35 \) → \( x = 3.5 \), which is correct.

So the steps are:

1. Multiply the second equation \( 3x - y = 11 \) by -5 to get \( -15x + 5y = -55 \)

2. Add this to the first equation \( 5x - 5y = 20 \):

\( 5x - 5y + (-15x + 5y) = 20 + (-55) \)

\( -10x = -35 \)

3. Solve for \( x \): \( x = \frac{35}{10} = \frac{7}{2} \)

4. Substitute \( x = \frac{7}{2} \) into \( 3x - y = 11 \):

\( 3\times\frac{7}{2} - y = 11 \)

\( \frac{21}{2} - y = 11 \)

\( -y = 11 - \frac{21}{2} = \frac{22 - 21}{2} = \frac{1}{2} \)

\( y = -\frac{1}{2} \)

So the solution is \( x = \frac{7}{2} \), \( y = -\frac{1}{2} \)

Answer:

The solution to the system is \( x = \frac{7}{2} \) (or \( 3.5 \)) and \( y = -\frac{1}{2} \) (or \( -0.5 \))