To solve the given system of equations using the elimination method, we start with the two equations:
- \( 13x + y = 3 \)
- \( 2y = -23x + 10 \)
First, let's rearrange equation (2) to solve for \(y\):
\[ 2y = -23x + 10 \]
To isolate \(y\), divide the entire equation by 2:
\[ y = -\frac{23}{2}x + 5 \]
Now we have:
- \( 13x + y = 3 \)
- \( y = -\frac{23}{2}x + 5 \)
Next, we will substitute equation (2) into equation (1) to eliminate \(y\):
Substituting for \(y\) in equation (1):
\[ 13x + \left(-\frac{23}{2}x + 5\right) = 3 \]
Now, simplify the equation:
\[ 13x - \frac{23}{2}x + 5 = 3 \]
To combine the terms involving \( x \), let's convert \( 13x \) into a fraction:
\[ 13x = \frac{26}{2}x \]
Now the equation becomes:
\[ \frac{26}{2}x - \frac{23}{2}x + 5 = 3 \]
Combine the \( x \) terms:
\[ \frac{26 - 23}{2} x + 5 = 3 \]
Simplifying that gives:
\[ \frac{3}{2} x + 5 = 3 \]
Next, isolate \( \frac{3}{2} x \):
\[ \frac{3}{2} x = 3 - 5 \]
\[ \frac{3}{2} x = -2 \]
To solve for \( x \), multiply both sides by the reciprocal of \( \frac{3}{2} \):
\[ x = -2 \cdot \frac{2}{3} \]
\[ x = -\frac{4}{3} \]
Now that we have \( x \), we substitute it back into one of the original equations to find \( y \). Using equation (1):
\[ 13\left(-\frac{4}{3}\right) + y = 3 \]
Calculating \( 13(-\frac{4}{3}) \):
\[ -\frac{52}{3} + y = 3 \]
Next, convert \(3\) into a fraction with a denominator of 3:
\[ 3 = \frac{9}{3} \]
Now the equation looks like this:
\[ -\frac{52}{3} + y = \frac{9}{3} \]
Isolating \(y\):
\[ y = \frac{9}{3} + \frac{52}{3} \]
\[ y = \frac{9 + 52}{3} \]
\[ y = \frac{61}{3} \]
Thus, the solution to the system of equations is:
\[ x = -\frac{4}{3} \]
\[ y = \frac{61}{3} \]
We can summarize the solution as:
Solution: \[ \left( x, y \right) = \left( -\frac{4}{3}, \frac{61}{3} \right) \]
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