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?

A = 6x + 3y = 14

B = 3x + 2y = 10

1 answer

To eliminate one of the variables when adding the equations A and B, we can manipulate the coefficients of either \( x \) or \( y \).

Let's start with the two equations:

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

To eliminate the variable \( y \), we can make the coefficients of \( y \) in both equations equal. The coefficient of \( y \) in Equation A is 3, and in Equation B, it is 2.

To make these coefficients equal, we can multiply the second equation (Equation B) by 3:

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

This results in:

\[ 9x + 6y = 30 \quad \text{(Modified Equation B)} \]

Now, we also need to modify Equation A to introduce a coefficient of 6 for \( y \):

Equation A remains: \[ 6x + 3y = 14 \]

Now, we can rewrite the two equations for addition:

  1. \( 6x + 3y = 14 \)
  2. \( 9x + 6y = 30 \)

Now, we can eliminate \( y \). To do this, we can multiply the first equation by 2:

\[ 2(6x + 3y) = 2(14) \] This gives us: \[ 12x + 6y = 28 \quad \text{(Modified Equation A)} \]

Now we can add the modified Equation A and modified Equation B:

  1. \( 12x + 6y = 28 \)
  2. \( 9x + 6y = 30 \)

At this point, since both equations now have \( 6y \), when we subtract one from the other, the \( 6y \) terms will cancel out.

Thus, to eliminate \( y \) when Tatum adds the two equations, she should multiply the second equation by \( 3 \).

Answer: 3