Question

Solve the system using elimination.

2x+4y=10


3x+5y=11


(1 point)
Responses

(4,−3)
open paren 4 comma negative 3 close paren

(3,1)
open paren 3 comma 1 close paren

(−3,4)
open paren negative 3 comma 4 close paren

no solution
no solution

Answer 1

To solve the system of equations using elimination, we can manipulate the equations so that when we add or subtract them, one of the variables will be eliminated.

The two equations are:

1) \( 2x + 4y = 10 \)
2) \( 3x + 5y = 11 \)

Let's multiply the first equation by 3 and the second equation by 2 so that the coefficients of \(x\) in both equations become equal:

\[
3(2x + 4y) = 3(10) \implies 6x + 12y = 30 \quad (3)
\]

\[
2(3x + 5y) = 2(11) \implies 6x + 10y = 22 \quad (4)
\]

Now we have:

3) \( 6x + 12y = 30 \)
4) \( 6x + 10y = 22 \)

Next, we subtract equation (4) from equation (3):

\[
(6x + 12y) - (6x + 10y) = 30 - 22
\]

This simplifies to:

\[
12y - 10y = 30 - 22
\]

\[
2y = 8
\]

Now, divide by 2:

\[
y = 4
\]

Now we can substitute \(y = 4\) back into one of the original equations to find \(x\). We'll use the first equation:

\[
2x + 4(4) = 10
\]

\[
2x + 16 = 10
\]

Subtract 16 from both sides:

\[
2x = 10 - 16
\]

\[
2x = -6
\]

Now, divide by 2:

\[
x = -3
\]

So the solution to the system of equations is:

\[
(x, y) = (-3, 4)
\]

The correct response is \((-3, 4)\).