Question

if you were to solve the following system by substitution, what would be the best variable to solve for and from what equation?
$2x + 6y = 9$
$3x - 12y = 15$

a. $y$, in the first equation
b. $x$, in the second equation
c. $x$, in the first equation
d. $y$, in the second equation

Explanation:

To determine the best variable to solve for in the substitution method, we look for a variable with a coefficient that is a factor of the constant term (or makes isolation easy).

• For the first equation \(2x + 6y = 9\), solving for \(x\): \(2x=9 - 6y\) \(\Rightarrow x=\frac{9 - 6y}{2}=\frac{9}{2}-3y\) (involves fractions). Solving for \(y\): \(6y = 9 - 2x\) \(\Rightarrow y=\frac{9 - 2x}{6}=\frac{3}{2}-\frac{1}{3}x\) (also fractions).
• For the second equation \(3x - 12y = 15\), solving for \(x\): \(3x=15 + 12y\) \(\Rightarrow x = 5 + 4y\) (no fractions, simple division by 3). Solving for \(y\): \(-12y=15 - 3x\) \(\Rightarrow y=\frac{3x - 15}{12}=\frac{x - 5}{4}\) (fractions, but solving for \(x\) here is simpler).

So solving for \(x\) in the second equation is easiest (no fractions, straightforward).

Answer:

B. \(x\), in the second equation