To eliminate the \( x \)-terms when adding the two equations, we can look at the coefficients of the \( x \) terms in both equations. The first equation is:
\[ 6x - 3y = 3 \quad \text{(Equation 1)} \]
The second equation is:
\[ -2x + 6y = 14 \quad \text{(Equation 2)} \]
Eliminate \( x \)-terms
- The coefficient of \( x \) in Equation 1 is \( 6 \).
- The coefficient of \( x \) in Equation 2 is \( -2 \).
To eliminate \( x \)-terms, we need these coefficients to be the same (but opposite in sign). We can multiply Equation 2 by \( 3 \) because:
\[ 3(-2x) = -6x \]
Now, if we multiply Equation 2 by \( 3 \), we get:
\[ 3(-2x + 6y) = 3(14) \]
This results in:
\[ -6x + 18y = 42 \quad \text{(Modified Equation 2)} \]
Now we can add the modified Equation 2 to Equation 1:
\[ (6x - 3y) + (-6x + 18y) = 3 + 42 \]
The \( x \) terms will cancel out, allowing us to solve for \( y \).
Thus, you would multiply the second equation by \( 3 \).
Eliminate \( y \)-terms
To eliminate the \( y \)-terms when adding the two equations, we look at their coefficients. The first equation has a \( -3y \) and the second equation has a \( 6y \).
- To eliminate \( y \)-terms, we can find a common multiple of the coefficients \( -3 \) and \( 6 \). The least common multiple is \( 6 \).
- To do this, we can multiply the first equation by \( 2 \) because:
\[ 2(-3y) = -6y \]
Now, if we multiply Equation 1 by \( 2 \), we get:
\[ 2(6x - 3y) = 2(3) \]
This results in:
\[ 12x - 6y = 6 \quad \text{(Modified Equation 1)} \]
Now we can add the modified Equation 1 to Equation 2 to eliminate \( y \):
\[ (12x - 6y) + (-2x + 6y) = 6 + 14 \]
The \( y \) terms will cancel out, allowing us to solve for \( x \).
Thus, you would multiply the first equation by \( 2 \).
Summary
- Multiply the second equation by \( 3 \) to eliminate \( x \)-terms.
- Multiply the first equation by \( 2 \) to eliminate \( y \)-terms.