Question

Tatum wants to solve the following system of equations by using the elimination method. What integer should she multiply the second equation by to eliminate a variable when you add the equations? 6x+3y=14 3x+2y=10 (1 point) Responses −2 negative 2 2 2 −12 negative 1 half 6 6 Skip to navigation

Answer 1

To eliminate a variable using the elimination method, Tatum can choose to eliminate either \(x\) or \(y\) by making the coefficients of one variable equal across both equations.

Given the system of equations:

1. \(6x + 3y = 14\) (Equation 1)
2. \(3x + 2y = 10\) (Equation 2)

We can choose to eliminate \(y\). The coefficient of \(y\) in Equation 1 is \(3\) and in Equation 2 is \(2\). To make them equal, we can multiply the second equation by \(1.5\) (or \(\frac{3}{2}\)) so that the coefficients become \(3\):

- Multiply Equation 2 by \(1.5\):

\[
1.5(3x + 2y) = 1.5(10)
\]

This results in:

\[
4.5x + 3y = 15
\]

Now we can add the new Equation 2 to Equation 1:

1. \(6x + 3y = 14\)
2. \(4.5x + 3y = 15\)

However, since \(1.5\) isn't listed as an option, we can instead look for a method to eliminate \(x\).

To eliminate \(x\), we observe the coefficients 6 and 3: if we multiply the second equation by \(2\) (which will yield \(6\) for the \(x\) term), we can set them up for elimination:

- Multiply the second equation (Equation 2) by \(2\):

\[
2(3x + 2y) = 2(10)
\]

This results in:

\[
6x + 4y = 20
\]

Now, comparing the scaled equations:

1. \(6x + 3y = 14\)
2. \(6x + 4y = 20\)

When we subtract Equation 1 from Equation 2:

\[
(6x + 4y) - (6x + 3y) = 20 - 14
\]

This gives us:

\[
y = 6
\]

So, to eliminate the \(x\) variable when adding the equations, Tatum should multiply the second equation by \(2\).

The answer is \(2\).