To solve the given system of equations using elimination, we'll follow these steps:
Given Equations:
- \(-5x - 3y = -13\) (Equation 1)
- \(4x + 2y = 12\) (Equation 2)
Step 1: Make the coefficients of \(y\) the same.
To eliminate \(y\), we can multiply Equation 1 by \(2\) (to get the coefficient of \(y\) as \(-6\)) and Equation 2 by \(3\) (to get the coefficient of \(y\) as \(6\)). This gives us:
First equation multiplied by 2: \[ 2(-5x - 3y) = 2(-13) \ -10x - 6y = -26 \quad (Equation 1') \]
Second equation multiplied by 3: \[ 3(4x + 2y) = 3(12) \ 12x + 6y = 36 \quad (Equation 2') \]
Step 2: Write the equivalent equations.
(a) First Equation: \[ -10x - 6y = -26 \]
(b) Second Equation: \[ 12x + 6y = 36 \]
Step 3: Add the two equations.
Now we will add Equation 1' and Equation 2': \[ (-10x - 6y) + (12x + 6y) = -26 + 36 \] This simplifies to: \[ (2x + 0y) = 10 \ 2x = 10 \]
Now, divide both sides by \(2\): \[ x = 5 \]
Step 4: Substitute \(x\) back to find \(y\).
Use the value of \(x\) in either original equation (let's use Equation 2): \[ 4(5) + 2y = 12 \ 20 + 2y = 12 \ 2y = 12 - 20 \ 2y = -8 \ y = -4 \]
Final Solutions:
- \(x = 5\)
- \(y = -4\)
LCM Values:
(a) The LCM of the x-values is: \[ \text{LCM}(5) = 5 \]
(b) The LCM of the y-values is: \[ \text{LCM}(-4) = 4 \]
Final Result Summary:
- (b) First equation: \(-10x - 6y = -26\)
- (b) Second equation: \(12x + 6y = 36\)
- (c) The operation used to eliminate: addition
- (d) When solving,
- \(x = 5\)
- \(y = -4\)
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