One Variable Already Isolated

Solve the system: $$\begin{cases} y = 3x - 4 \\ 2x + y = 6 \end{cases}$$

Identify the isolated variable: The first equation already has $y$ isolated: $y = 3x - 4$ = Ready to substitute

Substitute into the second equation: Replace $y$ with $(3x - 4)$ in $2x + y = 6$: $2x + (3x - 4) = 6$ = $2x + 3x - 4 = 6$

Solve for $x$: $5x - 4 = 6$ $5x = 10$ $x = 2$ = $x = 2$

Back-substitute to find $y$: $y = 3(2) - 4 = 6 - 4 = 2$ = $y = 2$

Verify the solution: Check in both equations: $y = 3(2) - 4 = 2$ ✓ $2(2) + 2 = 6$ ✓ = Solution verified

Answer: $(2, 2)$

Isolating a Variable First

Solve the system: $$\begin{cases} x + 2y = 8 \\ 3x - y = 3 \end{cases}$$

Choose which variable to isolate: In equation 1, $x$ has coefficient 1, so isolate $x$: $x = 8 - 2y$ = $x = 8 - 2y$

Substitute into the second equation: Replace $x$ in $3x - y = 3$: $3(8 - 2y) - y = 3$ = $24 - 6y - y = 3$

Solve for $y$: $24 - 7y = 3$ $-7y = -21$ $y = 3$ = $y = 3$

Back-substitute to find $x$: $x = 8 - 2(3) = 8 - 6 = 2$ = $x = 2$

Verify the solution: Check: $2 + 2(3) = 8$ ✓ $3(2) - 3 = 3$ ✓ = Solution verified

Answer: $(2, 3)$

System with Fractions

Solve the system: $$\begin{cases} 2x + 3y = 12 \\ x = y + 1 \end{cases}$$

Identify the isolated variable: The second equation has $x$ isolated: $x = y + 1$ = Ready to substitute

Substitute into the first equation: Replace $x$ with $(y + 1)$ in $2x + 3y = 12$: $2(y + 1) + 3y = 12$ = $2y + 2 + 3y = 12$

Solve for $y$: $5y + 2 = 12$ $5y = 10$ $y = 2$ = $y = 2$

Back-substitute to find $x$: $x = 2 + 1 = 3$ = $x = 3$

Verify the solution: Check: $2(3) + 3(2) = 6 + 6 = 12$ ✓ $3 = 2 + 1$ ✓ = Solution verified

Answer: $(3, 2)$

Real-World Application: Ticket Sales

A theater sold 150 tickets. Adult tickets cost 12 dollars and child tickets cost 8 dollars. Total revenue was 1480 dollars. How many of each type were sold?

Define variables and write equations: Let $a$ = adult tickets, $c$ = child tickets Total tickets: $a + c = 150$ Total revenue: $12a + 8c = 1480$ = System created

Isolate one variable: From equation 1: $a = 150 - c$ = $a = 150 - c$

Substitute into revenue equation: $12(150 - c) + 8c = 1480$ $1800 - 12c + 8c = 1480$ = $1800 - 4c = 1480$

Solve for $c$: $-4c = -320$ $c = 80$ = 80 child tickets

Find adult tickets: $a = 150 - 80 = 70$ = 70 adult tickets

Verify: $70 + 80 = 150$ ✓ $12(70) + 8(80) = 840 + 640 = 1480$ ✓ = Solution verified

Answer: 70 adult tickets and 80 child tickets

Mistake: Forgetting to distribute when substituting

Why: When substituting an expression like $(3x - 4)$, the parentheses indicate the entire expression replaces the variable.

Correct: Always use parentheses and distribute: $2(3x - 4) = 6x - 8$, not $2 \cdot 3x - 4 = 6x - 4$

Mistake: Substituting back into the wrong equation

Why: If you substitute back into the equation you used for substitution, you might get a trivial identity instead of a value.

Correct: After finding one variable, substitute into the OTHER original equation or the isolation equation.

Mistake: Sign errors when isolating variables

Why: Moving terms across the equals sign requires changing signs.

Correct: From $x + 2y = 8$, we get $x = 8 - 2y$ (not $x = 8 + 2y$)

Mistake: Not verifying the solution in both equations

Why: The solution must satisfy both equations simultaneously. Checking only one equation may miss errors.

Correct: Always substitute $(x, y)$ into BOTH original equations to verify.

Business Break-Even Analysis

Companies use systems of equations to find where revenue equals costs.

If cost $C = 500 + 10x$ (500 dollars fixed plus 10 dollars per item) and revenue $R = 25x$ (25 dollars per item), break-even is when $C = R$: $500 + 10x = 25x$, so $x = 33.3$ items.

Mixture Problems

Scientists and pharmacists use systems to create solutions with specific concentrations.

To make 100 mL of 30% acid solution from 20% and 50% solutions: Let $x$ = mL of 20%, $y$ = mL of 50%. Then $x + y = 100$ and $0.20x + 0.50y = 30$.

Q: When should I use substitution instead of elimination?

A: Use substitution when one equation already has a variable isolated (like $y = 2x + 3$) or when a variable has coefficient 1 or -1. Use elimination when variables have coefficients that are easy to match or cancel.

Q: What if I get something like $0 = 5$?

A: If you get a false statement like $0 = 5$, the system has no solution. The lines are parallel and never intersect.

Q: What if I get something like $0 = 0$?

A: If you get a true statement like $0 = 0$ with no variables, the system has infinitely many solutions. The equations represent the same line.