To solve the given system of equations using elimination, we first rewrite the equations in standard form.
The given equations are:
- \( 4x = 12 - 8y \)
- \( 5x + 3y = 6 \)
We can rearrange the first equation to standard form:
\[ 4x + 8y = 12 \]
Now we have the system of equations:
- \( 4x + 8y = 12 \)
- \( 5x + 3y = 6 \)
Next, we can eliminate one of the variables. Let's eliminate \( y \) by first making the coefficients of \( y \) in both equations equal.
To do this, we can multiply the entire first equation by 3 and the entire second equation by 8:
-
\( 3(4x + 8y) = 3(12) \)
- Which simplifies to: \( 12x + 24y = 36 \)
-
\( 8(5x + 3y) = 8(6) \)
- Which simplifies to: \( 40x + 24y = 48 \)
Now, we have a new system:
- \( 12x + 24y = 36 \)
- \( 40x + 24y = 48 \)
Next, we subtract the first equation from the second equation:
\[ (40x + 24y) - (12x + 24y) = 48 - 36 \]
This simplifies to:
\[ 28x = 12 \]
Now, solve for \( x \):
\[ x = \frac{12}{28} = \frac{3}{7} \]
Now that we have \( x \), we can substitute \( x \) back into one of the original equations to find \( y \). We'll use the first equation:
\[ 4x + 8y = 12 \]
Substituting \( x = \frac{3}{7} \):
\[ 4\left(\frac{3}{7}\right) + 8y = 12 \]
This simplifies to:
\[ \frac{12}{7} + 8y = 12 \]
Subtract \(\frac{12}{7}\) from both sides:
\[ 8y = 12 - \frac{12}{7} \]
To combine the terms on the right, convert 12 into a fraction with a denominator of 7:
\[ 12 = \frac{84}{7} \]
So:
\[ 8y = \frac{84}{7} - \frac{12}{7} = \frac{72}{7} \]
Now, divide both sides by 8:
\[ y = \frac{72}{7} \cdot \frac{1}{8} = \frac{72}{56} = \frac{9}{7} \]
Thus, the solution to the system of equations is:
\[ x = \frac{3}{7}, \quad y = \frac{9}{7} \]
Conclusion
The correct choice is:
A. The solution to the system is \( x = \frac{3}{7} \), \( y = \frac{9}{7} \).