Related Formula
A point (x₀, y₀)$(x_0, y_0)$ is a solution to a system of equations if substituting (x₀, y₀)$(x_0, y_0)$ into each equation yields a true identity.
Core Logic
Substitute x = -(4)/(7)$x = -\frac{4}{7}$ and y = -(1)/(7)$y = -\frac{1}{7}$ into Ax + 5y = -3$Ax + 5y = -3$:
A(-(4)/(7)) + 5(-(1)/(7)) = -3$$A\left(-\frac{4}{7}\right) + 5\left(-\frac{1}{7}\right) = -3$$
-(4A)/(7) - (5)/(7) = -3$$-\frac{4A}{7} - \frac{5}{7} = -3$$
Multiply by 7:
-4A - 5 = -21 -4A = -16 A = 4$$-4A - 5 = -21 \implies -4A = -16 \implies A = 4$$
Step 1: Verification in Second Equation
Check A = 4$A = 4$ in -8x + Ay = 4$-8x + Ay = 4$:
-8(-(4)/(7)) + 4(-(1)/(7)) = (32 - 4)/(7) = (28)/(7) = 4 (Verified)$$-8\left(-\frac{4}{7}\right) + 4\left(-\frac{1}{7}\right) = \frac{32 - 4}{7} = \frac{28}{7} = 4 \quad \text{(Verified)}$$
Thus, A = 4$A = 4$ (Option C).
Pattern Recognition
Given solution point (x, y)$(x, y)$ arrow$\rightarrow$ substitute directly into linear equation to solve for unknown parameter A$A$.
Chapter Mix
SAT Math: Heart of Algebra