QUESTION IMAGE
Question
- solve the following system of linear equations by elimination.\
\
what is the value of $y$?\
\bigcirc\\ y = 1.5\
\bigcirc\\ y = -1.5\
\bigcirc\\ y = 3\
\bigcirc\\ y = -3
Step1: Multiply equations to eliminate \(x\)
Multiply the first equation \(2x - 5y = 1\) by \(3\) and the second equation \(-3x + 2y = -18\) by \(2\) to make the coefficients of \(x\) opposites.
First equation: \(3(2x - 5y) = 3(1)\) gives \(6x - 15y = 3\).
Second equation: \(2(-3x + 2y) = 2(-18)\) gives \(-6x + 4y = -36\).
Step2: Add the two new equations
Add \(6x - 15y = 3\) and \(-6x + 4y = -36\) to eliminate \(x\):
\((6x - 6x) + (-15y + 4y) = 3 + (-36)\)
Simplify: \(-11y = -33\).
Step3: Solve for \(y\)
Divide both sides by \(-11\): \(y = \frac{-33}{-11} = 3\).
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\(y = 3\) (corresponding to the option " \(y = 3\) ")