Graphing Linear Inequalities and Systems of Inequalities: Algebra 1 Math Worksheet Answer Key – PDF Download

Each solution gives the boundary line, the line type, the test point and the shading decision, so you can see where a graph went wrong. Compare the described region with your own drawing, and confirm any point you chose by substituting it into every inequality.

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

Problem 1: Test a point

Substitute $x=2$ and $y=1$: $1<3(2)$, which is $1<6$. True.

Yes, $(2, 1)$ is a solution.

Problem 2: Points on the boundary

$2(3)+3(2)=12$, and $12\ge 12$ is true because $\ge$ allows equality. Yes.
$2(1)+3(2)=8$, and $8\ge 12$ is false. No.

Problem 3: Solid or dashed

$>$: dashed.
$\le$: solid.
$<$: dashed.
$\ge$: solid.

A line is solid only when the symbol includes equality.

Problem 4: Graph y > x + 2

The boundary $y=x+2$ has y-intercept $2$ and slope $1$. The symbol $>$ makes it dashed.

Test $(0, 0)$: $0>0+2$ is false, so shade the side that does not contain the origin: above the line.

The shaded region is above the dashed line.

Problem 5: Graph y ≤ −2x + 3

The boundary $y=-2x+3$ passes through $(0, 3)$ and $(1, 1)$. The symbol $\le$ makes it solid.

Test $(0, 0)$: $0\le 3$ is true, so shade the side with the origin: below the line.

Solid line, shaded below.

Problem 6: Graph with a fractional slope

The boundary passes through $(0, -1)$ and, using rise $1$ and run $2$, through $(2, 0)$ and $(4, 1)$. It is solid.

Test $(0, 0)$: $0\ge -1$ is true, so shade above the line, where the origin is.

Solid line, shaded above.

Problem 7: Standard form

Set $y=0$: $2x=6$, so $x=3$. Set $x=0$: $3y=6$, so $y=2$. The boundary joins $(3, 0)$ and $(0, 2)$ and is dashed.

Test $(0, 0)$: $0<6$ is true, so shade the side that contains the origin.

Dashed boundary with the origin side shaded.

Problem 8: Divide by a negative

Divide every term by $-3$ and reverse the symbol: $y\le -2x+3$.

The symbol $\le$ gives a solid line. The test point $(0, 0)$ gives $0\le 3$, true, so the shading is below the line.

Problem 9: Vertical and horizontal

$x\ge -2$: draw a solid vertical line through $x=-2$ and shade everything to its right.

$y<3$: draw a dashed horizontal line through $y=3$ and shade everything below it.

The graph of $x\ge -2$.

The graph of $y<3$.

Problem 10: Write from a graph: dashed

The line passes through $(0, 1)$ and $(2, 5)$, so the slope is $\frac{5-1}{2-0}=2$ and the equation is $y=2x+1$.

The line is dashed, so the symbol is strict. The shaded side is below the line, and the origin is shaded: $0<2(0)+1$ is true.

The inequality is $y<2x+1$.

Problem 11: Write from a graph: solid

The line passes through $(0, 2)$ and $(2, 0)$, so the slope is $\frac{0-2}{2-0}=-1$ and the equation is $y=-x+2$.

The line is solid and the shading is above it. Check a shaded point such as $(2, 2)$: $2\ge -2+2=0$ is true.

The inequality is $y\ge -x+2$.

Problem 12: Check points in a system

$(0, 0)$: $0>-1$ true; $0\le 5$ true. Solution.
$(3, 1)$: $1>2$ false. Not a solution.
$(2, 2)$: $2>1$ true; $2\le 3$ true. Solution.
$(4, 2)$: $2>3$ false. Not a solution.
$(-1, 4)$: $4>-2$ true; $4\le 6$ true. Solution.

$(0, 0)$, $(2, 2)$ and $(-1, 4)$ are solutions.

Problem 13: Graph a system

Both boundaries are dashed. $y=-x+4$ passes through $(0, 4)$ and $(4, 0)$; shade below. $y=2x-5$ passes through $(0, -5)$ and $(3, 1)$; shade above.

Set $-x+4=2x-5$: $9=3x$, so $x=3$ and $y=1$. The boundaries meet at $(3, 1)$.

Test the corner: $1<-3+4=1$ is false, so $(3, 1)$ is not a solution. The solution region is the overlap to the left of that point.

The overlap is the solution region.

Problem 14: A triangular region

$x\ge 0$ shades right of the y-axis and $y\ge 0$ shades above the x-axis. The line $x+y=6$ joins $(6, 0)$ and $(0, 6)$; shade below it. All three lines are solid.

The corners are $(0, 0)$, $(6, 0)$ and $(0, 6)$.

Test $(2, 3)$: $2\ge 0$, $3\ge 0$ and $2+3=5\le 6$ are all true, so it is a solution.

The feasible triangle.

Problem 15: A rectangular region

The inequalities form four solid boundaries: $x=-2$, $x=3$, $y=1$ and $y=4$.

The corners are $(-2, 1)$, $(3, 1)$, $(3, 4)$ and $(-2, 4)$.

The width is $3-(-2)=5$ and the height is $4-1=3$, so the area is $5\times 3=$ 15 square units.

The rectangular solution region.

Problem 16: No overlap

The boundary lines have the same slope $1$, so they are parallel. $y>x+3$ shades above the upper line, and $y<x-1$ shades below the lower line.

A solution would need $x+3<y<x-1$, which would make $x+3<x-1$, or $3<-1$. That is false.

The shaded regions never overlap, so there is no solution.

The two shaded regions are separated.

Problem 17: Concert budget

The total cost must be at most $\$150$: $15x+25y\le 150$.

$15(4)+25(3)=60+75=135\le 150$. Yes.
$15(5)+25(3)=75+75=150\le 150$. Yes, exactly the budget.
$15(2)+25(5)=30+125=155>150$. No.

Problem 18: Bake sale limits

At most 60 items: $x+y\le 60$. At least 20 cookies: $y\ge 20$. Counts cannot be negative: $x\ge 0$.

$(35, 25)$: $60\le 60$ and $25\ge 20$. Possible.
$(45, 20)$: $45+20=65>60$. Not possible.
$(30, 15)$: $15<20$. Not possible.
$(20, 20)$: $40\le 60$ and $20\ge 20$. Possible.

Possible: $(35, 25)$ and $(20, 20)$.

Problem 19: Yard work hours

Time: $2x+y\le 10$. Counts: $x\ge 0$ and $y\ge 0$.

The line $2x+y=10$ meets the axes at $(5, 0)$ and $(0, 10)$. The corners are $(0, 0)$, $(5, 0)$ and $(0, 10)$.

3 lawns and 4 walks: $2(3)+4=10\le 10$. Fits.

4 lawns and 3 walks: $2(4)+3=11>10$. Does not fit.

Problem 20: Lemonade goal

Earnings: $2x+3y\ge 60$. Cookie limit: $y\le 25$.

$(15, 10)$: $30+30=60\ge 60$ and $10\le 25$. Meets both.
$(10, 10)$: $20+30=50<60$. Fails.
$(30, 25)$: $60+75=135\ge 60$ and $25\le 25$. Meets both.

(15, 10) and (30, 25) work.

The shaded region meets both conditions.

Problem 21: Find the error

Mistake 1: the symbol $>$ is strict, so the line must be dashed.

Mistake 2: test $(0, 0)$: $3(0)-2(0)=0>6$ is false, so the origin side must not be shaded.

Boundary: $y=0$ gives $x=2$; $x=0$ gives $y=-3$. Test $(4, 0)$: $12>6$ is true, so shade the side containing $(4, 0)$, which is below and to the right of the line.

The correct graph.

Problem 22: Garden fence

The perimeter is $2x+2y\le 40$, which simplifies to $x+y\le 20$. With the minimums: $x\ge 8$ and $y\ge 4$.

Corners: $x=8,\ y=4$ gives $(8, 4)$. With $y=4$, $x+4=20$ gives $(16, 4)$. With $x=8$, $8+y=20$ gives $(8, 12)$.

Test $(10, 8)$: $10\ge 8$, $8\ge 4$ and $18\le 20$ are all true, so a 10 ft by 8 ft garden is allowed.

The feasible region has corners (8, 4), (16, 4) and (8, 12).

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