To use the elimination method to solve the system of equations:
- \( 6x + 3y = 14 \) (Equation 1)
- \( 3x + 2y = 10 \) (Equation 2)
we want to eliminate one of the variables by making the coefficients of either \( x \) or \( y \) the same in both equations.
Let's eliminate \( y \):
The coefficient of \( y \) in Equation 1 is \( 3 \), and the coefficient of \( y \) in Equation 2 is \( 2 \). To eliminate \( y \), we can multiply the second equation by \( 1.5 \) (or \( \frac{3}{2} \)). This way, the coefficient of \( y \) in Equation 2 will become \( 3 \), matching that of Equation 1.
Let's multiply Equation 2 by \( 1.5 \):
\[ 1.5(3x + 2y) = 1.5(10) \]
This simplifies to:
\[ 4.5x + 3y = 15 \]
Now, the system of equations becomes:
- \( 6x + 3y = 14 \)
- \( 4.5x + 3y = 15 \)
To eliminate \( y \), we can subtract the two equations:
\[ (6x + 3y) - (4.5x + 3y) = 14 - 15 \]
This results in:
\[ 1.5x = -1 \]
Solving for \( x \):
\[ x = \frac{-1}{1.5} = -\frac{2}{3} \]
If you are required to eliminate \( y \), you can multiply the second equation \( 3x + 2y = 10 \) by \( 3 \):
When you multiply the second equation by \( 3 \):
\[ 3(3x + 2y = 10) \] gives: \[ 9x + 6y = 30 \]
Now we can proceed to add/subtract the equations accordingly.
The integer Tatum should multiply the second equation by to eliminate \( y \) when you add the equations is:
\[ \text{multiply } 3 \text{ to the second equation.} \]