Systems of equations and inequalities

Chapter 3 · 6 lessons · 316 questions

Unit 1: How many solutions does a system have?

Key points

A system of two lines has one solution, no solution, or infinitely many solutions.
Write both equations as $ax + by = c$ . Then compare three ratios: $\dfrac{a_1}{a_2}$ , $\dfrac{b_1}{b_2}$ and $\dfrac{c_1}{c_2}$ .
If $\dfrac{a_1}{a_2}$ and $\dfrac{b_1}{b_2}$ are different, the lines cross once: one solution.
If $\dfrac{a_1}{a_2}$ and $\dfrac{b_1}{b_2}$ are equal but $\dfrac{c_1}{c_2}$ is different, the lines are parallel: no solution.
If all three ratios are equal, the two equations are the same line: infinitely many solutions.

See it solved

Example

$2x + 3y = 12$
$6y = 30 - 4x$

How many solutions does the given system of equations have?

AZero
BExactly one
CExactly two
DInfinitely many

Worked by hand

The plan

Write both equations in standard form
Evaluate the ratios

Desmos: Desmos is quicker here: type both equations and look. Zoom out before you decide that two lines are parallel.

Step 1: Write both equations in standard form
- The first equation is already in standard form: $2x + 3y = 12$ .
- In the second equation, add $4x$ to both sides: $4x + 6y = 30$ .
Step 2: Evaluate the ratios
- $a_1 = 2$ , $b_1 = 3$ and $c_1 = 12$ . $a_2 = 4$ , $b_2 = 6$ and $c_2 = 30$ .
- $\dfrac{a_1}{a_2} = \dfrac{2}{4} = \dfrac{1}{2}$ .
- $\dfrac{b_1}{b_2} = \dfrac{3}{6} = \dfrac{1}{2}$ .
- $\dfrac{c_1}{c_2} = \dfrac{12}{30} = \dfrac{2}{5}$ .
- The first two ratios are equal, and the third is different.
- So the lines are parallel and never meet. There is no solution: choice A.

The correct answer choice is A: the system has

0

solutions.

The same example in Desmos

new graph
Desmos shows: The first line.
Desmos shows: A second line, beside the first.
zoom out
Desmos shows: The lines stay the same distance apart. They never cross, so there is no solution.

Desmos expression list: 2x+3y=12; 6y=30-4x — Type

Desmos graph of 2x+3y=12; 6y=30-4x — Look

Made with Desmos

Open Desmos

Your turn

$x - 2y = 5$
$3x - 6y = 15$

How many solutions does the given system of equations have?

AZero
BExactly one
CExactly two
DInfinitely many

Show the answerHide the answer

Answer

The correct answer choice is D: all three ratios equal $\dfrac{1}{3}$ , so the system has infinitely many solutions.

Step 1: Write both equations in standard form
- Both equations are already in standard form.
Step 2: Evaluate the ratios
- $a_1 = 1$ , $b_1 = -2$ and $c_1 = 5$ . $a_2 = 3$ , $b_2 = -6$ and $c_2 = 15$ .
- $\dfrac{a_1}{a_2} = \dfrac{1}{3}$ .
- $\dfrac{b_1}{b_2} = \dfrac{-2}{-6} = \dfrac{1}{3}$ .
- $\dfrac{c_1}{c_2} = \dfrac{5}{15} = \dfrac{1}{3}$ .
- All three ratios are equal, so the two equations are the same line.
- Every point on that line is a solution: infinitely many.

Traps

Stopping after two ratios. Equal $\dfrac{a_1}{a_2}$ and $\dfrac{b_1}{b_2}$ only says the lines have the same slope. Check $\dfrac{c_1}{c_2}$ too. Here it is different, so ‘Infinitely many’ (choice D) is a wrong answer.
Comparing before both equations are in the same form. In $6y = 30 - 4x$ the $x$ -term is on the right. Reading $a_2 = -4$ gives $\dfrac{2}{-4} = -\dfrac{1}{2}$ , which does not match $\dfrac{1}{2}$ . That leads to ‘Exactly one’ (choice B), a wrong answer.

Practice this lesson32 questions

Unit 2: Finding the constant: no solution or infinitely many

Key points

No solution: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}$ , but $\dfrac{c_1}{c_2}$ is different.
Infinitely many solutions: all three ratios are equal.
To find an unknown constant, set two ratios equal and cross-multiply.
Then test the third ratio, to be sure which case you have.

See it solved

Example

$3x + ky = 7$
$6x + 10y = 5$

In the given system of equations, $k$ is a constant. The system has no solution. What is the value of $k$ ?

Worked by hand

The plan

Equate the ratios
Test the third ratio

Desmos: By hand is quicker and exact. If you are comfortable with sliders, a slider for $k$ is a check: at the right value the two lines are parallel.

Step 1: Equate the ratios
- No solution means $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}$ .
- $a_1 = 3$ and $b_1 = k$ . $a_2 = 6$ and $b_2 = 10$ .
- $\dfrac{3}{6} = \dfrac{k}{10}$ .
- Cross-multiply: $6 \cdot k = 3 \cdot 10$ .
- $3 \cdot 10 = 30$ , so $6k = 30$ .
- $k = \dfrac{30}{6} = 5$ .
Step 2: Test the third ratio
- $\dfrac{a_1}{a_2} = \dfrac{3}{6} = \dfrac{1}{2}$ .
- $\dfrac{c_1}{c_2} = \dfrac{7}{5}$ .
- $\dfrac{7}{5}$ is not $\dfrac{1}{2}$ . So the lines are parallel and apart: no solution.
- $k = 5$ works.

The correct answer is

5

The same example in Desmos

new graph
Desmos shows: Desmos offers to add a slider for $k$ . Accept it: a row $k = 1$ appears.
Desmos shows: The second line. The two lines cross.
type 5 into the k slider's row
Desmos shows: The two lines are parallel and apart. No solution, so $k = 5$ .

Desmos expression list: 3x+ky=7; k=5; 6x+10y=5 — Type

Desmos graph of 3x+ky=7; k=5; 6x+10y=5 — Look

Made with Desmos

Open Desmos

Your turn

$4x - 6y = 10$
$kx - 9y = 15$

In the given system of equations, $k$ is a constant. The system has infinitely many solutions. What is the value of $k$ ?

Show the answerHide the answer

Answer

The correct answer is $6$ .

Step 1: Equate the ratios
- Infinitely many solutions means all three ratios are equal.
- $\dfrac{b_1}{b_2} = \dfrac{-6}{-9} = \dfrac{2}{3}$ .
- $\dfrac{c_1}{c_2} = \dfrac{10}{15} = \dfrac{2}{3}$ . These two match.
- So $\dfrac{a_1}{a_2} = \dfrac{4}{k}$ must be $\dfrac{2}{3}$ too: $\dfrac{4}{k} = \dfrac{2}{3}$ .
- Cross-multiply: $2 \cdot k = 4 \cdot 3$ .
- $4 \cdot 3 = 12$ , so $2k = 12$ .
- $k = \dfrac{12}{2} = 6$ .

Traps

Flipping one of the ratios. $\dfrac{3}{6} = \dfrac{k}{10}$ compares first equation over second both times. Writing $\dfrac{3}{6} = \dfrac{10}{k}$ gives $3k = 60$ , so $k = 20$ , a wrong answer.
Skipping the third ratio. Equal $\dfrac{a_1}{a_2}$ and $\dfrac{b_1}{b_2}$ fits ‘no solution’ and ‘infinitely many’. Only $\dfrac{c_1}{c_2}$ tells them apart. Here $\dfrac{7}{5}$ is not $\dfrac{1}{2}$ , so it is ‘no solution’.

Practice this lesson60 questions

Unit 3: Solving a system, and the add-or-subtract shortcut

Key points

The solution of a system is the point $(x, y)$ that makes both equations true. It is where the two lines cross.
Substitution: if one variable is already alone, plug its expression into the other equation.
Elimination: add or subtract the equations so that one variable disappears.
If the question asks for $x + y$ or $x - y$ , try adding or subtracting the equations first. You may get it in one step.

See it solved

Example

$3x + 2y = 16$
$x - 2y = 0$

The solution to the given system of equations is $(x, y)$ . What is the value of $y$ ?

Worked by hand

The plan

Determine the approach
Determine the value of $x$
Determine the value of $y$

Desmos: Desmos is quicker here: type both equations and click where the lines cross.

Step 1: Determine the approach
- The $y$ -terms are $+2y$ and $-2y$ . They are opposites.
- So add the two equations. The $y$ -terms will disappear: this is elimination.
Step 2: Determine the value of $x$
- Add the left sides: $3x + x = 4x$ and $2y - 2y = 0$ .
- Add the right sides: $16 + 0 = 16$ .
- $4x = 16$ , so $x = \dfrac{16}{4} = 4$ .
Step 3: Determine the value of $y$
- Plug $x = 4$ into $x - 2y = 0$ : $4 - 2y = 0$ .
- Add $2y$ to both sides: $4 = 2y$ .
- $y = \dfrac{4}{2} = 2$ .
- Check in the first equation: $3 \cdot 4 + 2 \cdot 2$ $= 12 + 4$ $= 16$ . It matches.

The correct answer is

2

The same example in Desmos

new graph
Desmos shows: The first line.
Desmos shows: The second line. The two lines cross once.
click the crossing
Desmos shows: $(4, 2)$ . So $x = 4$ and $y = 2$ .

Desmos expression list: 3x+2y=16; x-2y=0 — Type

Desmos graph of 3x+2y=16; x-2y=0, with (4, 2) labelled — Look

Made with Desmos

Open Desmos

Your turn

$5x + 3y = 31$
$3x + 5y = 25$

The solution to the given system of equations is $(x, y)$ . What is the value of $x + y$ ?

Show the answerHide the answer

Answer

The correct answer is $7$ .

Step 1: Determine the approach
- The question asks for $x + y$ , not for $x$ or $y$ alone.
- Try adding the two equations.
Step 2: Determine the value of $x + y$
- Add the left sides: $5x + 3x = 8x$ and $3y + 5y = 8y$ .
- Add the right sides: $31 + 25 = 56$ .
- $8x + 8y = 56$ .
- Divide both sides by $8$ : $x + y = \dfrac{56}{8} = 7$ .

Traps

Answering with the wrong variable. $x = 4$ is found first, but the question asks for $y$ . $4$ is a wrong answer.
Subtracting when the terms are opposites. $+2y$ and $-2y$ cancel when you add. Subtracting gives $2x + 4y = 16$ , and no variable is gone.

Practice this lesson104 questions

Unit 4: Solving a linear inequality

Key points

Solve an inequality like an equation: do the same thing to both sides.
One extra rule: when you multiply or divide both sides by a negative number, flip the sign.
$<$ and $>$ leave the boundary out. $\le$ and $\ge$ include it.
The answer is a range of values, not one number.

See it solved

Example

$-3x + 5 \le 20$

What is the least value of $x$ that satisfies the given inequality?

Worked by hand

The plan

Isolate the variable
Determine the least value

Desmos: By hand takes two lines. Desmos is a check: type the inequality and click the boundary. A solid line means the boundary is included.

Step 1: Isolate the variable
- Subtract $5$ from both sides: $-3x \le 20 - 5$ , so $-3x \le 15$ .
- Divide both sides by $-3$ . Dividing by a negative number flips the sign.
- $x \ge \dfrac{15}{-3}$ , so $x \ge -5$ .
Step 2: Determine the least value
- $x \ge -5$ means $x$ is $-5$ or more.
- The sign $\ge$ includes $-5$ itself, so the least value is $-5$ .
- Check $x = -5$ : $-3 \cdot (-5) + 5$ $= 15 + 5$ $= 20$ . Is $20$ at most $20$ ? Yes.
- Check $x = -6$ : $-3 \cdot (-6) + 5$ $= 18 + 5$ $= 23$ . Is $23$ at most $20$ ? No. So $-6$ is not allowed.

The correct answer is

-5

The same example in Desmos

new graph
Desmos shows: Shading to the right of a solid vertical line.
click where the line meets the x-axis
Desmos shows: $(-5, 0)$ . The shading is on its right and the line is solid, so $x \ge -5$ .

Desmos expression list: -3x+5le20 — Type

Desmos graph of -3x+5le20, with (−5, 0) labelled — Look

Made with Desmos

Open Desmos

Your turn

$7 - 2x > 1$

What is the greatest integer value of $x$ that satisfies the given inequality?

Show the answerHide the answer

Answer

The correct answer is $2$ .

Step 1: Isolate the variable
- Subtract $7$ from both sides: $-2x > 1 - 7$ , so $-2x > -6$ .
- Divide both sides by $-2$ and flip the sign: $x < \dfrac{-6}{-2}$ , so $x < 3$ .
Step 2: Determine the greatest value
- $x < 3$ does not include $3$ .
- The greatest integer less than $3$ is $2$ .
- Check $x = 2$ : $7 - 2 \cdot 2$ $= 7 - 4$ $= 3$ . Is $3$ greater than $1$ ? Yes.

Traps

Not flipping the sign. Dividing $-3x \le 15$ by $-3$ without flipping gives $x \le -5$ . That says $-5$ is the greatest value and there is no least value, a wrong answer.
Flipping at the wrong time. Adding or subtracting a number never flips the sign. Dividing by a positive number never flips it. Only multiplying or dividing by a negative number does.

Practice this lesson36 questions

Unit 5: Systems of inequalities: points and regions

Key points

A point is a solution of a system only if it makes every inequality true.
To test a point, plug in its $x$ and its $y$ . Stop at the first inequality that is false.
On a graph, $y >$ shades above the line and $y <$ shades below it.
A dashed line ( $<$ or $>$ ) is not part of the solution. A solid line ( $\le$ or $\ge$ ) is.
In Desmos, the solutions are where the two shadings overlap.

See it solved

Example

$y > 2x - 1$
$y \le -x + 5$

Which point $(x, y)$ is a solution to the given system of inequalities?

A $(1, 1)$
B $(1, 3)$
C $(2, 4)$
D $(3, 2)$

Worked by hand

The plan

Test each point in the first inequality
Test the remaining points in the second inequality

Desmos: Desmos is quicker here: type both inequalities, then type the points. The solution is the point inside the darker overlap.

Step 1: Test each point in the first inequality
- The first inequality is $y > 2x - 1$ .
- A, $(1, 1)$ : $2 \cdot 1 - 1 = 1$ . Is $1$ greater than $1$ ? No, they are equal. Rule out A.
- B, $(1, 3)$ : $2 \cdot 1 - 1 = 1$ . Is $3$ greater than $1$ ? Yes.
- C, $(2, 4)$ : $2 \cdot 2 - 1 = 3$ . Is $4$ greater than $3$ ? Yes.
- D, $(3, 2)$ : $2 \cdot 3 - 1 = 5$ . Is $2$ greater than $5$ ? No. Rule out D.
Step 2: Test the remaining points in the second inequality
- The second inequality is $y \le -x + 5$ .
- B, $(1, 3)$ : $-1 + 5 = 4$ . Is $3$ at most $4$ ? Yes.
- C, $(2, 4)$ : $-2 + 5 = 3$ . Is $4$ at most $3$ ? No. Rule out C.
- Only B makes both inequalities true.

The correct answer choice is B:

(1, 3)

The same example in Desmos

new graph
Desmos shows: Shading above a dashed line.
Desmos shows: Shading below a solid line. The darker overlap is the solution set.
Desmos shows: Four points. Only $(1, 3)$ is inside the darker overlap. $(1, 1)$ sits on the dashed line, which does not count.

Desmos expression list: y>2x-1; yle-x+5; (1,1); (1,3); (2,4); (3,2) — Type

Desmos graph of y>2x-1; yle-x+5; (1,1); (1,3); (2,4); (3,2), with (1, 3) labelled — Look

Made with Desmos

Open Desmos

Your turn

$y \ge x + 1$
$y < 3x$

The point $(2, y)$ is a solution to the given system of inequalities. What is the greatest integer value of $y$ ?

Show the answerHide the answer

Answer

The correct answer is $5$ .

Step 1: Plug the point into both inequalities
- In $y \ge x + 1$ , put $x = 2$ : $y \ge 2 + 1$ , so $y \ge 3$ .
- In $y < 3x$ , put $x = 2$ : $y < 3 \cdot 2$ , so $y < 6$ .
Step 2: Chain the two inequalities
- $3 \le y < 6$ .
- $6$ is not included. The integers that work are $3$ , $4$ and $5$ .
- The greatest is $5$ .

Traps

Testing only one inequality. $(2, 4)$ passes $y > 2x - 1$ but fails $y \le -x + 5$ . A solution has to pass both. $(2, 4)$ is choice C, a wrong answer.
Counting a point on a dashed line. $(1, 1)$ gives $1$ on both sides of $y > 2x - 1$ . Equal is not greater, so the point is not a solution. That is choice A, a wrong answer.

Practice this lesson52 questions

Unit 6: No solution or infinitely many solutions

Key points

Simplify each side until it looks like (a number) $x$ + (a number).
Different $x$ -coefficients: exactly one solution.
The same $x$ -coefficients and different constants: no solution.
The same $x$ -coefficients and the same constants: infinitely many solutions.
A side with no $x$ -term has an $x$ -coefficient of $0$ .

See it solved

Example

$4(x + 3) - x$ $= kx + 7$

In the given equation, $k$ is a constant. The equation has no solution. What is the value of $k$ ?

Worked by hand

The plan

Evaluate the coefficients and constants
Equate the coefficients

Desmos: With a constant to find, equate the coefficients by hand. Desmos is a check: graph each side as a line. Parallel lines mean no solution.

Step 1: Evaluate the coefficients and constants
- Multiply out the left side: $4 \cdot x = 4x$ and $4 \cdot 3 = 12$ .
- So the left side is $4x + 12 - x$ .
- $4x - x = 3x$ . The left side is $3x + 12$ .
- The right side is $kx + 7$ .
Step 2: Equate the coefficients
- No solution means the $x$ -coefficients are equal and the constants are different.
- The $x$ -coefficients are $3$ and $k$ . So $k = 3$ .
- The constants are $12$ and $7$ . They are different, so there is no solution.
- Check: $3x + 12 = 3x + 7$ . Subtract $3x$ from both sides: $12 = 7$ , which is false for every $x$ .

The correct answer is

3

The same example in Desmos

new graph
Desmos shows: The left side, as a line.
Desmos shows: The right side with $k = 3$ : a parallel line.
zoom out
Desmos shows: The lines never cross. So with $k = 3$ there is no solution.

Desmos expression list: y=4(x+3)-x; y=3x+7 — Type

Desmos graph of y=4(x+3)-x; y=3x+7 — Look

Made with Desmos

Open Desmos

Your turn

$2(3x - 5) + 4$ $= 6x - 6$

How many solutions does the given equation have?

AZero
BExactly one
CExactly two
DInfinitely many

Show the answerHide the answer

Answer

The correct answer choice is D: both sides are $6x - 6$ , so there are infinitely many solutions.

Step 1: Evaluate the coefficients and constants
- Multiply out the left side: $2 \cdot 3x = 6x$ and $2 \cdot (-5) = -10$ .
- So the left side is $6x - 10 + 4$ .
- $-10 + 4 = -6$ . The left side is $6x - 6$ .
- The right side is $6x - 6$ .
Step 2: Equate the coefficients
- The $x$ -coefficients are both $6$ . The constants are both $-6$ .
- The two sides are the same, so every value of $x$ works: infinitely many solutions.

Traps

Forgetting the lone $-x$ . $4(x + 3) - x$ is $3x + 12$ , not $4x + 12$ . Stopping at $4x + 12$ gives $k = 4$ , a wrong answer.
Mixing up the two cases. Equal coefficients with different constants give no solution. Equal coefficients with equal constants give infinitely many. If the right side were $3x + 12$ , every $x$ would work.

Practice this lesson32 questions

Typing tips for Desmos

In the keys lines, → means the right-arrow key.
Typing / starts a fraction, and everything after it goes into the denominator until you press →. So 42/120*100 typed straight through becomes 42 ÷ (120 × 100). Type 42/120 → *100 instead.
Desmos solves equations in x (and y) only. An equation in t, such as 3t + 7 = 22, gives an error until you retype it with x.

Start practicing

Practice by lesson

Lesson 1, How many solutions does a system have?, 32 questions
Lesson 2, Finding the constant: no solution or infinitely many, 60 questions
Lesson 3, Solving a system, and the add-or-subtract shortcut, 104 questions
Lesson 4, Solving a linear inequality, 36 questions
Lesson 5, Systems of inequalities: points and regions, 52 questions
Lesson 6, No solution or infinitely many solutions, 32 questions

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Next chapter · Chapter 4Linear word problems

Desmos

Next set

1Unit 1: How many solutions does a system have?

Key points

See it solved

Worked by hand

Step 1: Write both equations in standard form

Step 2: Evaluate the ratios

The same example in Desmos

Your turn

Step 1: Write both equations in standard form

Step 2: Evaluate the ratios

Traps

2Unit 2: Finding the constant: no solution or infinitely many

Key points

See it solved

Worked by hand

Step 1: Equate the ratios

Step 2: Test the third ratio

The same example in Desmos

Your turn

Step 1: Equate the ratios

Traps

3Unit 3: Solving a system, and the add-or-subtract shortcut

Key points

See it solved

Worked by hand

Step 1: Determine the approach

Step 2: Determine the value of xxx

Step 3: Determine the value of yyy

The same example in Desmos

Your turn

Step 1: Determine the approach

Step 2: Determine the value of x+yx + yx+y

Traps

4Unit 4: Solving a linear inequality

Key points

See it solved

Worked by hand

Step 1: Isolate the variable

Step 2: Determine the least value

The same example in Desmos

Your turn

Step 1: Isolate the variable

Step 2: Determine the greatest value

Traps

5Unit 5: Systems of inequalities: points and regions

Key points

See it solved

Worked by hand

Step 1: Test each point in the first inequality

Step 2: Test the remaining points in the second inequality

The same example in Desmos

Your turn

Step 1: Plug the point into both inequalities

Step 2: Chain the two inequalities

Traps

6Unit 6: No solution or infinitely many solutions

Key points

See it solved

Worked by hand

Step 1: Evaluate the coefficients and constants

Step 2: Equate the coefficients

The same example in Desmos

Your turn

Step 1: Evaluate the coefficients and constants

Step 2: Equate the coefficients

Traps

Practice by lesson

Mixed practice: every lesson

Easy set 1

800-level question

What tripped you up

Unit 1: How many solutions does a system have?

Unit 2: Finding the constant: no solution or infinitely many

Unit 3: Solving a system, and the add-or-subtract shortcut

Step 2: Determine the value of $x$

Step 3: Determine the value of $y$

Step 2: Determine the value of $x + y$

Unit 4: Solving a linear inequality

Unit 5: Systems of inequalities: points and regions

Unit 6: No solution or infinitely many solutions