When One Variable is Already Isolated

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

Identify the isolated variable: The first equation already has $y$ isolated: $y = 3x - 2$ = $y = 3x - 2$

Substitute into the other equation: Replace $y$ with $(3x - 2)$ in the second equation: $2x + (3x - 2) = 8$ = $2x + 3x - 2 = 8$

Solve for $x$: $5x - 2 = 8$ $5x = 10$ $x = 2$ = $x = 2$

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

Verify the solution: Check in second equation: $2(2) + 4 = 4 + 4 = 8$ ✓ = Solution verified!

Answer: The solution is $(2, 4)$

Isolating a Variable First

Solve the system: $$\begin{cases} x + y = 7 \\ 2x - 3y = -6 \end{cases}$$

Choose which variable to isolate: From the first equation, isolate $x$: $x = 7 - y$ = $x = 7 - y$

Substitute into the other equation: Replace $x$ with $(7 - y)$ in the second equation: $2(7 - y) - 3y = -6$ = $2(7 - y) - 3y = -6$

Distribute and simplify: $14 - 2y - 3y = -6$ $14 - 5y = -6$ = $14 - 5y = -6$

Solve for $y$: $-5y = -6 - 14$ $-5y = -20$ $y = 4$ = $y = 4$

Back-substitute to find $x$: $x = 7 - 4 = 3$ = $x = 3$

Verify in both equations: Equation 1: $3 + 4 = 7$ ✓ Equation 2: $2(3) - 3(4) = 6 - 12 = -6$ ✓ = Both equations satisfied!

Answer: The solution is $(3, 4)$

System with Fractions in the Solution

Solve the system: $$\begin{cases} y = 2x + 1 \\ 4x - y = 5 \end{cases}$$

Identify the isolated variable: $y$ is already isolated in the first equation = $y = 2x + 1$

Substitute into the second equation: $4x - (2x + 1) = 5$ = $4x - 2x - 1 = 5$

Solve for $x$: $2x - 1 = 5$ $2x = 6$ $x = 3$ = $x = 3$

Find $y$: $y = 2(3) + 1 = 7$ = $y = 7$

Verify: Check: $4(3) - 7 = 12 - 7 = 5$ ✓ = Verified!

Answer: The solution is $(3, 7)$

Choosing the Best Variable to Isolate

Solve the system: $$\begin{cases} 3x + y = 10 \\ 5x - 2y = -4 \end{cases}$$

Choose the easiest variable to isolate: In the first equation, $y$ has coefficient 1, so isolate $y$: $y = 10 - 3x$ = $y = 10 - 3x$

Substitute into the second equation: $5x - 2(10 - 3x) = -4$ = $5x - 2(10 - 3x) = -4$

Distribute and combine: $5x - 20 + 6x = -4$ $11x - 20 = -4$ = $11x - 20 = -4$

Solve for $x$: $11x = 16$ $x = \frac{16}{11}$ = $x = \frac{16}{11}$

Find $y$: $y = 10 - 3 \cdot \frac{16}{11} = 10 - \frac{48}{11} = \frac{110 - 48}{11} = \frac{62}{11}$ = $y = \frac{62}{11}$

Verify: Check first equation: $3 \cdot \frac{16}{11} + \frac{62}{11} = \frac{48 + 62}{11} = \frac{110}{11} = 10$ ✓ = Solution verified!

Answer: The solution is $\left(\frac{16}{11}, \frac{62}{11}\right)$

Mistake: Forgetting to distribute when substituting

Why: When you substitute an expression like $(3x - 2)$ and there's a coefficient in front, you must distribute.

Correct: If you have $2(3x - 2)$, distribute to get $6x - 4$, not $6x - 2$.

Mistake: Substituting into the same equation

Why: If you solve equation 1 for $y$ and substitute back into equation 1, you'll just get $0 = 0$ (always true) and learn nothing.

Correct: Always substitute into the OTHER equation to create a new equation with one variable.

Mistake: Sign errors when distributing negatives

Why: Subtracting an expression like $-(2x + 3)$ means multiplying each term by $-1$.

Correct: $-(2x + 3) = -2x - 3$, not $-2x + 3$.

Mistake: Not checking the solution in BOTH equations

Why: A solution must satisfy both equations. Checking only one equation doesn't guarantee correctness.

Correct: Always substitute your answer into both original equations to verify.

Comparing Phone Plans

Phone companies offer different plans. The substitution method helps find when two plans cost the same.

Plan A costs 20 euros monthly plus 0.10 euros per minute. Plan B costs 30 euros monthly plus 0.05 euros per minute. When are they equal?

Mixing Solutions in Chemistry

Scientists often need to mix solutions of different concentrations to get a desired result.

A chemist needs 100 mL of a 30% acid solution. They have 20% and 50% solutions available.

Budget Planning

Businesses use systems of equations to balance costs and revenue.

A company sells basic and premium products. If they sell 50 items and make 2000 euros, with basic at 30 euros and premium at 50 euros, how many of each did they sell?

Q: When should I use substitution vs. elimination?

A: Use substitution when one variable is already isolated or has a coefficient of 1. Use elimination when both equations have similar coefficients that can be easily canceled.

Q: What if I get $0 = 0$ or a contradiction like $0 = 5$?

A: If you get $0 = 0$, the system has infinitely many solutions (the lines are the same). If you get something impossible like $0 = 5$, there's no solution (parallel lines).

Q: Does it matter which variable I isolate first?

A: Mathematically, no - you'll get the same answer either way. Practically, choose the variable that's easiest to isolate (smallest coefficient, already partially isolated).