Find the solution to the system of equations.

Question

You can use the interactive graph below to find the solution.

\[\begin{cases}
8x-4y=16
\\\\
8x+4y=16
\end{cases}\]
\[x= \]
\[y= \]

Answer 1

To solve the system of equations given by:

\[ \begin{cases} 8x - 4y = 16 \quad \text{(1)} \ 8x + 4y = 16 \quad \text{(2)} \end{cases} \]

let's first manipulate these equations.

1. From equation (1): \[ 8x - 4y = 16 \] We can solve for \(y\): \[ -4y = 16 - 8x \] \[ 4y = 8x - 16 \] \[ y = 2x - 4 \quad \text{(3)} \]

2. From equation (2): \[ 8x + 4y = 16 \] Similarly, we can solve for \(y\): \[ 4y = 16 - 8x \] \[ y = 4 - 2x \quad \text{(4)} \]

Now we have two expressions for \(y\) from equations (3) and (4): \[ y = 2x - 4 \quad \text{(3)} \] \[ y = 4 - 2x \quad \text{(4)} \]

Setting these two expressions for \(y\) equal to each other: \[ 2x - 4 = 4 - 2x \] Now, solve for \(x\): \[ 2x + 2x = 4 + 4 \] \[ 4x = 8 \] \[ x = 2 \]

Now substituting \(x = 2\) into either equation (3) or (4) to find \(y\). Let's use equation (3): \[ y = 2(2) - 4 \] \[ y = 4 - 4 \] \[ y = 0 \]

So, the solution to the system of equations is: \[ x = 2, \quad y = 0 \]

Final answer: \[ x = 2 \] \[ y = 0 \]