Algebra 1 Linear Inequalities in Two Variables and Systems Worksheet Solutions (PDF)

Algebra 1 math worksheet solutions: Linear Inequalities in Two Variables and Systems – Maths PDF

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This linear inequalities two variables answers for Algebra 1 covers linear inequalities in two variables and systems (Common Core A-REI.D.12, A-CED.A.3), with a free PDF to print.

Linear inequalities two variables answers (Algebra 1) – first page of the printable PDF

Here are full solutions for the inequalities in two variables worksheet. Read each step, not just the bold answer, and compare it with your own method. Redo any problem you missed, and note which kind of mistake you made so you can avoid it on the test.

Algebra 1 · Common Core State Standards: A-REI.D.12, A-CED.A.3

Linear inequalities two variables answers: exercises 1 to 10

Exercise 1 – Test a point

Substitute $x = 2$, $y = 1$: $1 < 2 + 2$, which is $1 < 4$.

Yes, it is a solution

Exercise 2 – Not a solution

Substitute: $-2 \ge 2(0) + 3$, which is $-2 \ge 3$, false.

No, it is not a solution

Exercise 3 – Solid or dashed

Strict inequalities (<, >) use dashed lines. Inequalities with ≤ or ≥ use solid lines.

(a) solid; (b) dashed; (c) dashed

Exercise 4 – Graph y > x + 1

Dashed line, shade above.

Boundary $y = x + 1$ is dashed. Test $(0, 0)$: $0 > 1$ is false, so shade the side without the origin, above the line.

Dashed line through (0, 1) with slope 1; shade above

Exercise 5 – Graph with a solid line

Solid line through the intercepts.

Intercepts: $(1, 0)$ and $(0, 4)$. The line is solid. Test $(0, 0)$: $0 \ge 4$ is false.

So shade the side that does not contain the origin.

Solid line through (1, 0) and (0, 4); shade above the line, away from the origin

Exercise 6 – Vertical boundary

Shade right of the line.

The boundary is the vertical line $x = 2$, solid. Solutions have $x$ at least 2, so shade to the right.

Solid vertical line at x = 2; shade right

Exercise 7 – Horizontal boundary

Shade below the dashed line.

The boundary $y = 3$ is dashed. Solutions have $y$ less than 3, so shade below.

Dashed horizontal line at y = 3; shade below

Exercise 8 – Write from a graph

The boundary passes through $(0, -2)$ with slope 1, so it is $y = x – 2$. It is dashed, and the shading is above.

$y > x – 2$

Exercise 9 – Standard form inequality

Solid line through the intercepts.

Intercepts: $(6, 0)$ and $(0, 4)$. Draw a solid line. Test $(0, 0)$: $0 \le 12$ is true, so shade the side with the origin.

Solid line through (6, 0) and (0, 4); shade toward the origin

Exercise 10 – Negative coefficient

Dashed line, shade below.

Solve for $y$: $-y > -3x + 3$, so $y < 3x - 3$ (reverse the symbol).

Dashed line through $(0, -3)$ with slope 3. Shade below.

Dashed line $y = 3x – 3$; shade below

Exercises 11 and up

Exercise 11 – Which points work?

$(0, 0): 0 \le 4$ true. $(3, 1): 3 + 2 = 5 \le 4$ false. $(-2, 5): -2 + 10 = 8 \le 4$ false.

$(0, 0)$ only

Exercise 12 – System of two inequalities

The overlap is the solution.

Graph the solid line $y = \frac{1}{2}x – 2$ and shade above. Graph the dashed line $y = -x + 4$ and shade below. The solution is the overlap.

The lines meet where $\frac{1}{2}x – 2 = -x + 4$, so $1.5x = 6$, at $x = 4$, $y = 0$.

Overlap region with corner (4, 0)

Exercise 13 – Triangular region

The triangle is the solution region.

The region is the triangle with corners $(0, 0)$, $(5, 0)$, $(0, 5)$.

Check $(2, 2)$: $2 \ge 0$, $2 \ge 0$, $2 + 2 = 4 \le 5$, all true.

Triangle with corners (0, 0), (5, 0), (0, 5); yes

Exercise 14 – Check a system

First: $2 > 3 – 2$, which is $2 > 1$, true. Second: $2 \le 2(3)$, which is $2 \le 6$, true.

Yes

Exercise 15 – Write a system from a graph

The solid boundary through $(0, -3)$ with slope 1 has shading above: $y \ge x – 3$.

The dashed horizontal boundary has shading below: $y < 2$.

$y \ge x – 3$ and $y < 2$

Exercise 16 – Lunch budget

The point (3, 5) is on the boundary line.

Total cost: $5x + 3y \le 30$.

For $(3, 5)$: $5(3) + 3(5) = 30 \le 30$, which is true.

$5x + 3y \le 30$; yes, exactly $30

Exercise 17 – Two jobs

Hours: $x + y \le 12$. Pay: $10x + 8y \ge 90$.

$(6, 5)$: $11 \le 12$ and $60 + 40 = 100 \ge 90$, so yes.

$(4, 4)$: $40 + 32 = 72 \ge 90$ is false.

(6, 5) works; (4, 4) does not

Exercise 18 – Whole-number combinations

For each $x$ from 0 to 6, count the whole numbers $y$ with $3y \le 12 – 2x$.

$x = 0$: $y = 0, 1, 2, 3, 4$ gives 5
$x = 1$: $y \le 3.33$ gives 4
$x = 2$: $y \le 2.67$ gives 3
$x = 3$: $y \le 2$ gives 3
$x = 4$: $y \le 1.33$ gives 2
$x = 5$: $y \le 0.67$ gives 1
$x = 6$: $y = 0$ gives 1

Total: $5 + 4 + 3 + 3 + 2 + 1 + 1 = 19$.

19 pairs

Exercise 19 – Truck load

$25x + 40y \le 280$.

$(4, 4)$: $100 + 160 = 260 \le 280$, yes.

$(6, 4)$: $150 + 160 = 310 \le 280$ is false, no.

(4, 4) yes; (6, 4) no

Exercise 20 – School store

$(6, 4)$: $6 + 4 = 10 \ge 10$ true; $6 + 12 = 18 \le 24$ true. Yes.

$(2, 8)$: $2 + 8 = 10 \ge 10$ true; $2 + 24 = 26 \le 24$ false. No.

(6, 4) yes; (2, 8) no

Exercise 21 – Corner of the region

Subtract the first equation from the second: $2n = 14$, so $n = 7$.

Then $p = 10 – 7 = 3$.

$(3, 7)$: 3 pencils and 7 notebooks

Key facts behind this linear inequalities two variables answers

Key facts

A solution of an inequality in two variables is an ordered pair that makes it true.
The boundary line is solid for ≤ and ≥, and dashed for < and >.
Test a point not on the line to decide which side to shade.
For y > mx + b, shade above; for y < mx + b, shade below.
The solution of a system is the overlap of the shaded regions.

How to use this linear inequalities two variables answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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