Example

Solving {4x3y=6,  15y20x=30}\{4x - 3y = 6,\; 15y - 20x = -30\} by Substitution

Solve the system {4x3y=615y20x=30\left\{\begin{array}{l} 4x - 3y = 6 \\ 15y - 20x = -30 \end{array}\right. using the substitution method.

Step 1 — Solve one equation for one variable. In the first equation, solve for xx. Add 3y3y to both sides:

4x=3y+64x = 3y + 6

Divide both sides by 44:

x=34y+32x = \frac{3}{4}y + \frac{3}{2}

Step 2 — Substitute into the other equation. Replace xx in the second equation with 34y+32\frac{3}{4}y + \frac{3}{2}:

15y20(34y+32)=3015y - 20\left(\frac{3}{4}y + \frac{3}{2}\right) = -30

Step 3 — Solve the resulting one-variable equation. Distribute 20-20 across the parentheses:

15y15y30=3015y - 15y - 30 = -30

Combine the like terms 15y15y=015y - 15y = 0:

030=300 - 30 = -30

0=00 = 0

Because 0=00 = 0 is a true statement — and the variable has been completely eliminated — the system is consistent and the equations are dependent. The two equations represent the same line, so the system has infinitely many solutions.

This example demonstrates what happens during substitution when the two equations in a system are actually the same line in disguise: every variable term cancels out and the result is a universally true numerical statement. Recognizing 0=00 = 0 as a signal of dependent equations is the algebraic counterpart of seeing two coincident lines on a graph.

0

1

Updated 2026-04-21

Contributors are:

Who are from:

Tags

OpenStax

Elementary Algebra @ OpenStax

Ch.5 Systems of Linear Equations - Elementary Algebra @ OpenStax

Algebra

Math

Prealgebra

Related
Learn After