ACT Math · Algebra · free lesson

Systems of Linear Equations

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The topic in one minute

Press Play: the idea and a worked example, explained aloud (about 1:19).

Systems of Linear Equations

The solution is where the lines cross: $(3, 5)$ .

1 / 7

Read what is said

A system is two equations in the same two letters. Its solution is one pair that makes both true, which is the point where the two lines cross. The ACT asks about this often.
Substitution comes first. The first equation says y is the same as two x minus one, so put that in parentheses in place of y in the other equation. Now only x is left to solve for.
Elimination makes one letter cancel. Multiply the first equation by two so its v term becomes plus four v, the opposite of minus four v. Then add, and v disappears.
Two lines can cross once, never meet, or lie on top of each other. Try substitution on parallel lines and the letters cancel, leaving a false statement. That means no solution.
Now an example. Here x is already alone in the first equation, so write two y plus one in parentheses in place of x in the second equation, and solve for y.
Here is the trap. Going back for x gives a second number, but the question asked for y. The other letter is exactly the wrong option the ACT is sure to offer, so read what is asked before you stop.
Remember: get one letter alone and substitute, or line up the equations and eliminate. Then go back for the other letter, check, and answer what is asked. Now practice it with new numbers.

The idea

Two equations at once

A system of linear equations is two equations in the same two letters, such as $y = 2x - 1$ and $3x + 2y = 19$ . A solution of the system is one pair $(x, y)$ that makes both equations true at the same time.

Each equation alone has a whole line of solutions. A pair that works in both must lie on both lines, so it is the point where the two lines cross. Watch the two lines draw themselves in and meet.

Definition.

The solution of a system of two linear equations is the point where their graphs cross. Its two numbers make both equations true.

Substitution

Example.

The first equation says that $y$ is the same thing as $2x - 1$ . So in the second equation write $2x - 1$ in place of $y$ , inside parentheses. That leaves one equation in $x$ alone:

\begin{aligned} 3x + 2(2x - 1) &= 19 \\ x &= 3 \\ y &= 2x - 1 \\ &= 2 \times 3 - 1 \\ &= 5 \end{aligned}

The solution is $({3}, {5})$ , the black point of the picture. Check it in the equation not used for the last step:

\begin{aligned} L &= 3x + 2y \\ &= 3 \times 3 + 2 \times 5 \\ &= 19 \end{aligned}

The check gives $19$ , the right side of the second equation.

Rule.

Substitution: get one letter alone in one equation; put what it equals, in parentheses, into the other equation and solve; then go back for the other letter.

Elimination

If two equations are true, their sum is true too, and so is either one multiplied by a number. Elimination uses this to make one letter cancel. Take $3u + 2v = 13$ and $5u - 4v = 7$ . Multiply the first equation by 2, so that its $2v$ becomes $4v$ , the opposite of $-4v$ :

\begin{aligned} P &= 2(3u + 2v) \\ &= 6u + 4v \\ Q &= 2 \times 13 \\ &= 26 \end{aligned}

Now add $6u + 4v = 26$ to $5u - 4v = 7$ . The $v$ terms cancel, and the sum is an equation in $u$ alone:

\begin{aligned} S &= P + 5u - 4v \\ &= 6u + 4v + 5u - 4v \\ &= 11u \\ S &= Q + 7 \\ 11u &= 26 + 7 \\ u &= 3 \\ 3u + 2v &= 13 \\ 3 \times 3 + 2v &= 13 \\ v &= 2 \end{aligned}

The solution is $u = 3$ , $v = 2$ . Check in the second equation:

\begin{aligned} L &= 5u - 4v \\ &= 5 \times 3 - 4 \times 2 \\ &= 7 \end{aligned}

Rule.

Elimination: line the equations up; multiply one or both so that one letter has opposite numbers in front; add; solve; go back for the other letter. Use substitution when a letter is already alone, elimination when both equations look like $ax + by = c$ .

When the question wants a combination. The ACT may ask for $x + y$ rather than $x$ or $y$ . In $4x + 3y = 26$ and $3x + 4y = 23$ , adding gives the same number of each letter, and dividing by 7 gives $x + y$ at once:

\begin{aligned} T &= 26 + 23 \\ &= 49 \\ s &= \frac{T}{7} \\ &= \frac{49}{7} \\ &= 7 \end{aligned}

So $x + y = 7$ , and neither letter had to be found.

One solution, none, or infinitely many

Two straight lines can do only three things, so a system of two linear equations has only three possible numbers of solutions.

The two lines	Slopes and intercepts	Solutions
cross at one point	different slopes	exactly one
are parallel	same slope, different $y$ -intercepts	none
are the same line	same slope, same $y$ -intercept	infinitely many

Try to solve $2x + y = 5$ and $6x + 3y = 21$ by substitution. From the first, $y = 5 - 2x$ :

\begin{aligned} L &= 6x + 3(5 - 2x) \\ &= 15 \\ 6x + 3(5 - 2x) &= 21 \\ N &= \text{no solution} \end{aligned}

The $x$ terms cancel and the left side is just $15$ , never 21: the engine reports no solution. The lines are parallel. Had the second equation been $6x + 3y = 15$ , the first one times 3, every $x$ would work:

\begin{aligned} 6x + 3(5 - 2x) &= 15 \\ M &= \text{infinitely many solutions} \end{aligned}

Drag the sliders. The blue line $y = 2x + 1$ stays put; the red line is $y = mx + c$ . The engine solves the system each time and says what it finds.

With any slope but 2 there is one crossing. With slope 2 the lines are parallel and there is no solution, unless $c$ is 1 as well: then the red line lies on the blue one.

Rule.

Write both equations as $ax + by = c$ and compare them number by number. If the $x$ and $y$ numbers are not in the same ratio, there is one solution. If they are, but the constants are not, there is none. If all three are in the same ratio, there are infinitely many.

A missing constant. For which $k$ has $3x + 2y = 7$ and $kx + 8y = 20$ no solution? The $y$ numbers are in the ratio 8 to 2, so the $x$ numbers must be too:

\begin{aligned} q &= \frac{8}{2} \\ &= 4 \\ \frac{k}{3} &= q \\ \frac{k}{3} &= 4 \\ k &= 12 \\ r &= \frac{20}{7} \end{aligned}

So $k$ is $12$ . The constants are in the ratio $\frac{20}{7}$ , not $4$ , so the lines are parallel, not the same line.

Systems from stories

A school sells 150 tickets to a play, adult tickets at 13 dollars and student tickets at 8 dollars, and takes in 1,400 dollars. With $a$ adult and $t$ student tickets, one equation counts tickets and one counts dollars: $a + t = 150$ and $13a + 8t = 1400$ . From the first, $t = 150 - a$ :

\begin{aligned} 13a + 8(150 - a) &= 1400 \\ a &= 40 \\ t &= 150 - a \\ &= 150 - 40 \\ &= 110 \end{aligned}

So $40$ adult and $110$ student tickets were sold. The two lines of the story cross at that point.

A system can also be written as one matrix equation, numbers in front of the letters in a matrix and the right sides in a column; topic 06 shows how.

Note.

On the ACT a system question asks for one letter, or a combination such as $x + y$ ; for the number of solutions; for a constant that gives no solution or infinitely many; or for a system set up from a story. Read what is asked before you stop: the wrong options include the other letter and a value from one equation alone.

Worked examples

Each example is one of the kinds of question the ACT asks about systems of linear equations. The numbers in the working are found by the engine.

Example.

1. Substitution, and reading what is asked. In the solution of the system $x = 2y + 1$ and $3x - 4y = 9$ , what is the value of $y$ ?

The first equation has $x$ alone. Put $2y + 1$ , in parentheses, in place of $x$ in the second, solve for $y$ , then go back for $x$ :

\begin{aligned} 3(2y + 1) - 4y &= 9 \\ y &= 3 \\ x &= 2y + 1 \\ &= 2 \times 3 + 1 \\ &= 7 \end{aligned}

Check in the second equation:

\begin{aligned} L &= 3x - 4y \\ &= 3 \times 7 - 4 \times 3 \\ &= 9 \end{aligned}

It gives $9$ , so the solution is $({7}, {3})$ . The question asks for $y$ , which is $3$ ; the value $7$ is $x$ , a wrong option the ACT is sure to offer.

Example.

2. Elimination with two multipliers. Solve $3x + 4y = 10$ and $2x - 5y = 22$ .

No letter is alone and no terms cancel yet. Make the $x$ terms equal: multiply the first equation by 2 and the second by 3. Each becomes an equation with $6x$ :

\begin{aligned} P &= 2(3x + 4y) \\ &= 6x + 8y \\ Q &= 3(2x - 5y) \\ &= 6x - 15y \end{aligned}

The right sides become $20$ and $66$ . Subtract the second new equation from the first, every term of it, and the $x$ terms cancel:

\begin{aligned} D &= P - Q \\ &= 6x + 8y - (6x - 15y) \\ &= 23y \\ D &= 2 \times 10 - 3 \times 22 \\ 23y &= 2 \times 10 - 3 \times 22 \\ y &= -2 \\ 3x + 4y &= 10 \\ 3x + 4 \times (-2) &= 10 \\ x &= 6 \end{aligned}

Check in the second equation:

\begin{aligned} L &= 2x - 5y \\ &= 2 \times 6 - 5 \times (-2) \\ &= 22 \end{aligned}

It gives $22$ , the right side, so the solution is $({6}, {-2})$ .

Warning.

When you subtract an equation, every term changes sign, the right side too: $-(-15y)$ is $+15y$ .

Example.

3. The crossing point of two graphs. The lines $y = -\frac{1}{2}x + 4$ and $y = x - 2$ are graphed below. Find the point where they meet.

At the crossing point the two lines have the same $y$ , so set the right sides equal and solve for $x$ :

\begin{aligned} \frac{-x}{2} + 4 &= x - 2 \\ x &= 4 \\ y &= x - 2 \\ &= 4 - 2 \\ &= 2 \end{aligned}

Check in the first equation:

\begin{aligned} L &= \frac{-x}{2} + 4 \\ &= -\frac{4}{2} + 4 \\ &= 2 \end{aligned}

Both lines give $y = 2$ at $x = 4$ : they meet at $({4}, {2})$ .

Example.

4. Constants for infinitely many solutions. For which values of $b$ and $c$ does the system $2x - 3y = 5$ and $6x + by = c$ have infinitely many solutions?

Infinitely many solutions means the two equations are one line, so the second must be the first with every term multiplied by one number $t$ . The $x$ terms give $t$ :

\begin{aligned} t &= \frac{6}{2} \\ &= 3 \\ b &= -3t \\ &= -3 \times 3 \\ &= -9 \\ c &= 5t \\ &= 5 \times 3 \\ &= 15 \end{aligned}

So $b$ is $-9$ and $c$ is $15$ : the second equation is $6x - {9}y = {15}$ , the first one times $3$ . With $b = -9$ but any other $c$ the lines would be parallel, with no solution.

Example.

5. A mixture. A chemist mixes a 20% acid solution with a 50% acid solution to make 30 liters of a 40% acid solution. How many liters of each does she use?

Let $p$ be the liters of the 20% solution and $q$ the liters of the 50% one. One equation counts liters; the other counts liters of acid.

	20% solution	50% solution	Mixture
Liters	$p$	$q$	30
Liters of acid	$0.2p$	$0.5q$	$0.4 \times 30$

So $p + q = 30$ and $0.2p + 0.5q = 0.4 \times 30$ . From the first, $q = 30 - p$ :

\begin{aligned} 0.2p + 0.5(30 - p) &= 0.4 \times 30 \\ p &= 10 \\ q &= 30 - p \\ &= 30 - 10 \\ &= 20 \end{aligned}

Check the acid:

\begin{aligned} A &= 0.2p + 0.5q \\ &= 0.2 \times 10 + 0.5 \times 20 \\ &= 12 \end{aligned}

She uses $10$ liters of the 20% solution and $20$ liters of the 50% solution, and the acid comes to $12$ liters, 40% of 30.

Example.

6. Two plans: when do they cost the same? A van rental costs 30 dollars plus 25 cents a mile with plan A, and 45 dollars plus 15 cents a mile with plan B. For how many miles do the two plans cost the same, and which is cheaper for a longer trip?

The cost in dollars for $d$ miles is $C = 0.25d + 30$ with plan A and $C = 0.15d + 45$ with plan B. Set the two costs equal:

\begin{aligned} 0.25d + 30 &= 0.15d + 45 \\ d &= 150 \end{aligned}

\begin{aligned} C &= 0.25d + 30 \\ &= 0.25 \times 150 + 30 \\ &= 67.5 \end{aligned}

At $150$ miles both plans cost $67.50$ dollars. Beyond that, plan B is cheaper, because its line is less steep: it adds less for each mile.

Practice

Answer ACT-style questions on systems of linear equations. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise systems of linear equations with new numbers