Question

Solving with Elimination Quick Check
4 of 54 of 5 Items






Question
Solve the system of equations.
3x−4y=40

5x−5y=55
(1 point)
Responses

(18, 7)
left parenthesis 18 comma 7 right parenthesis

(8,−3)
left parenthesis 8 comma negative 3 right parenthesis

(4,−7)
left parenthesis 4 comma negative 7 right parenthesis

(14, 3)

Answer 1

To solve the system of equations using elimination, let's first express the equations:

1. \( 3x - 4y = 40 \) (Equation 1)
2. \( 5x - 5y = 55 \) (Equation 2)

We want to eliminate one of the variables. We can do this by manipulating the equations.

Let's multiply Equation 1 by 5 and Equation 2 by 3 to align the coefficients of \(x\):

\[
5(3x - 4y) = 5(40) \implies 15x - 20y = 200 \quad \text{(Equation 3)}
\]
\[
3(5x - 5y) = 3(55) \implies 15x - 15y = 165 \quad \text{(Equation 4)}
\]

Now we can subtract Equation 4 from Equation 3:

\[
(15x - 20y) - (15x - 15y) = 200 - 165
\]
\[
-20y + 15y = 35
\]
\[
-5y = 35
\]
\[
y = -7
\]

Now that we found \(y\), we can substitute it back into one of the original equations to find \(x\). We'll use Equation 1:

\[
3x - 4(-7) = 40
\]
\[
3x + 28 = 40
\]
\[
3x = 40 - 28
\]
\[
3x = 12
\]
\[
x = 4
\]

So the solution to the system of equations is \((x, y) = (4, -7)\).

The final answer is:
\[
\boxed{(4, -7)}
\]