To solve the system of equations using the elimination method, let's analyze the provided equations:
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First equation: \[ 6x - 3y = 18 \]
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Second equation: \[ -3x - 9y = -93 \]
Question 20:
The goal of the elimination method is to eliminate one of the variables by combining the equations.
It looks like the first equation could be manipulated easily to eliminate \(x\) or \(y\). To eliminate \(x\), we want the coefficients of \(x\) in both equations to be equal. The coefficient of \(x\) in the first equation is \(6\) and in the second equation it's \(-3\). To make them equal, we could multiply the second equation by \(2\) because:
\[ -3x \times 2 = -6x \] This will allow us to eliminate \(x\) when we add the two equations together, giving us:
\(\text{Multiply the second equation by } -2\) does not help in achieving equal coefficients.
- Multiply the first equation by 3 results in \(18x - 9y = 54\), changing coefficients of \(y\).
- Multiply the first equation by -2 does not yield any useful results.
- Multiply the second equation by 2 gets us \(-6x - 18y = -186\), making it straightforward to combine.
- Multiply the second equation by -3 does not contribute positively.
Therefore, the correct choice for Question 20 is:
multiply the second equation by 2.
Question 21:
Now, using the equations given with the modifications from the eliminated variable, we proceed with the second equation multiplied by \(2\):
\[ 2(-3x - 9y) = 2(-93) \] This gives: \[ -6x - 18y = -186 \]
Now we add this to the modified first equation \(6x - 3y = 18\): \[ (6x - 3y) + (-6x - 18y) = 18 - 186 \] The \(6x\) and \(-6x\) will cancel: \[ -21y = -168 \] Now solving for \(y\): \[ y = \frac{-168}{-21} = 8 \]
Substituting \(y = 8\) back into one of the original equations to find \(x\), we use the first equation: \[ 6x - 3(8) = 18 \] \[ 6x - 24 = 18 \] \[ 6x = 18 + 24 \] \[ 6x = 42 \] \[ x = 7 \]
The value of \(x\) is:
7.
Summary
- Answer to Question 20: multiply the second equation by 2.
- Answer to Question 21: 7.
3 answers