Systems of Equations

We have learned how to solve linear systems graphically and algebraically. Now we develop a systematic way to classify systems and apply our skills to nonlinear systems and word problems.

Classification by Determinant

For a general system of two linear equations:

$\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}$

the key quantity is the determinant of the coefficient matrix.

⚡Theorem — Determinant Criterion

Let $D = a_1 b_2 - a_2 b_1$ . Then:

If $D \neq 0$ : the system has a unique solution. The lines intersect at exactly one point.
If $D = 0$ and $a_1 c_2 - a_2 c_1 \neq 0$ : the system has no solution. The lines are parallel.
If $D = 0$ and $a_1 c_2 - a_2 c_1 = 0$ : the system has infinitely many solutions. The lines are identical.

When $D \neq 0$ , the unique solution is given by Cramer’s rule:

$x = \dfrac{c_1 b_2 - c_2 b_1}{D}, \qquad y = \dfrac{a_1 c_2 - a_2 c_1}{D}$

The determinant gives a quick way to know what to expect before you begin solving.

Nonlinear Systems

Not all systems involve only lines. When a line meets a curve, we can still use substitution.

✎Example — Line Meets Parabola

Solve the system:

$\begin{cases} y = x^2 \\ y = x + 2 \end{cases}$

Solution. Set the right-hand sides equal:

$x^2 = x + 2$ $x^2 - x - 2 = 0$ $(x - 2)(x + 1) = 0$

So $x = 2$ or $x = -1$ .

When $x = 2$ : $y = 2^2 = 4$ . When $x = -1$ : $y = (-1)^2 = 1$ .

Solutions: $(2, 4)$ and $(-1, 1)$ .

A line and a parabola can intersect at 0, 1, or 2 points. Here we have two intersection points.

Word Problems

Many real-world situations translate naturally into systems of equations. The key skill is translating sentences into algebraic equations.

Age Problem

✎Example — Age Problem

Maria is 3 times as old as her brother. In 5 years, she will be twice as old as he will be. How old are they now?

Solution. Let $m$ = Maria’s current age and $b$ = her brother’s current age.

Translate the conditions:

“Maria is 3 times as old”: $m = 3b$
“In 5 years, twice as old”: $m + 5 = 2(b + 5)$

Solve by substitution. Replace $m$ with $3b$ in the second equation:

$3b + 5 = 2(b + 5)$ $3b + 5 = 2b + 10$ $b = 5$

Then $m = 3(5) = 15$ .

Answer: Maria is 15 years old and her brother is 5 years old.

Check: $15 = 3 \times 5$ ✓. In 5 years: $20 = 2 \times 10$ ✓.

Mixture Problem

✎Example — Mixture Problem

How much 20% salt solution and 50% salt solution must be mixed to obtain 300 g of a 30% salt solution?

Solution. Let $x$ = grams of 20% solution and $y$ = grams of 50% solution.

Total weight: $x + y = 300$

Total salt: $0.2x + 0.5y = 0.3 \times 300 = 90$

From equation 1: $x = 300 - y$ . Substitute:

$0.2(300 - y) + 0.5y = 90$ $60 - 0.2y + 0.5y = 90$ $0.3y = 30$ $y = 100$

Then $x = 300 - 100 = 200$ .

Answer: Mix 200 g of the 20% solution with 100 g of the 50% solution.

Check: $0.2(200) + 0.5(100) = 40 + 50 = 90 = 0.3(300)$ ✓.

✏Exercise — Practice

A movie theater sold 150 tickets. Adult tickets cost $12 and child tickets cost $7. The total revenue was $1400. How many adult tickets and how many child tickets were sold? Set up the system and solve.