Question

svlc algebra 1a - standard (15260)
solving systems: introduction to linear combinations
what is the solution (a, b) to this system of linear equations?
3a + 6b = 45
2a - 2b = -12
(-1, 7) (1, 7)
(-27, 6) (27, -6)

Explanation:

Step1: Simplify the first equation

Divide the first equation \(3a + 6b = 45\) by 3:
\(\frac{3a}{3}+\frac{6b}{3}=\frac{45}{3}\)
\(a + 2b = 15\) (Equation 1)

Step2: Simplify the second equation

Divide the second equation \(2a - 2b = -12\) by 2:
\(\frac{2a}{2}-\frac{2b}{2}=\frac{-12}{2}\)
\(a - b = -6\) (Equation 2)

Step3: Subtract Equation 2 from Equation 1

\((a + 2b)-(a - b)=15-(-6)\)
\(a + 2b - a + b = 15 + 6\)
\(3b = 21\)

Step4: Solve for \(b\)

Divide both sides by 3:
\(b=\frac{21}{3}=7\)

Step5: Substitute \(b = 7\) into Equation 2

\(a - 7 = -6\)
Add 7 to both sides:
\(a=-6 + 7 = 1\)

Answer:

(1, 7)