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Linear Equations and Literal Equations: Algebra 1 Math Worksheet Answer Key – PDF Download

    Linear Equations and Literal Equations – Algebra 1 math worksheet answer key

    Every solution below shows each step and ends with a check in the original equation. Compare your reasoning, not only your final answer: the standards ask students to justify each step. If your answer differs, find the first line where your work and the key disagree.

    Algebra 1Common Core: A-REI.A.1 · A-REI.B.3 · A-CED.A.1 · A-CED.A.4

    Problem 1: Two steps with a tape diagram

    Subtract 9 from both sides: \(4x=28\).

    Divide both sides by 4: \(x=7\).

    Check: \(4(7)+9=37\).

    3777779
    Four equal parts of size 7 and a part of size 9 make 37.

    Problem 2: Variables on both sides

    Subtract \(5x\) from both sides: \(2x-3=11\).

    Add 3: \(2x=14\). Divide by 2: \(x=7\).

    Check: \(7(7)-3=46\) and \(5(7)+11=46\).

    Problem 3: Distribute first

    Distribute: \(3x+12=2x+19\).

    Subtract \(2x\): \(x+12=19\). Subtract 12: \(x=7\).

    Check: \(3(11)=33\) and \(2(7)+19=33\).

    Problem 4: Fractional coefficient

    Add 5: \(\frac{2}{3}x=12\).

    Multiply by the reciprocal \(\frac{3}{2}\): \(x=12\cdot\frac{3}{2}\), so \(x=18\).

    Check: \(\frac{2}{3}(18)-5=12-5=7\).

    Problem 5: Fractions with a common denominator

    Multiply every term by the least common denominator 6: \(3x+2x=60\).

    Combine: \(5x=60\). Divide by 5: \(x=12\).

    Check: \(6+4=10\).

    Problem 6: Fractions on both sides

    Multiply both sides by 12: \(3(x+3)=4(2x-1)\).

    Distribute: \(3x+9=8x-4\). Subtract \(3x\): \(9=5x-4\). Add 4: \(13=5x\).

    Divide by 5: \(x=\frac{13}{5}\).

    Check: \(\frac{13/5+3}{4}=\frac{28/5}{4}=\frac{7}{5}\) and \(\frac{26/5-1}{3}=\frac{21/5}{3}=\frac{7}{5}\).

    Problem 7: Decimals on both sides

    Subtract \(0.4x\): \(1.2=0.3x-0.3\). Add 0.3: \(1.5=0.3x\).

    Divide by 0.3: \(x=5\).

    Check: \(0.4(5)+1.2=3.2\) and \(0.7(5)-0.3=3.2\).

    Problem 8: Give the reasons

    Step 2: distributive property. Step 3: combine like terms. Step 4: subtraction property of equality (subtract \(x\) from both sides). Step 5: addition property of equality (add 1 to both sides).

    Check: \(x=8\) gives \(2(5)+5=15\) and \(8+7=15\).

    2(x – 3) + 5 = x + 7Given2x – 6 + 5 = x + 7Distributive property2x – 1 = x + 7Combine like termsx – 1 = 7Subtraction property of equalityx = 8Addition property of equality
    Reasons for each step of \(2(x-3)+5=x+7\).

    Problem 9: Find the error

    Jamal multiplied only the first term. The distributive property gives \(3x-6=12\).

    Add 6: \(3x=18\). Divide by 3: \(x=6\).

    Check: \(3(6-2)=12\).

    Problem 10: No solution or not?

    Distribute: \(2x+6=2x+5\). Subtract \(2x\): \(6=5\).

    This is false for every \(x\), so the equation has no solution.

    Problem 11: Infinitely many solutions?

    Distribute and combine: \(4x-8+3=4x-5\), so \(4x-5=4x-5\).

    Both sides are identical, so the statement is always true: infinitely many solutions (all real numbers).

    Test \(x=0\): \(-5=-5\).

    Problem 12: Solve by graphing

    The lines cross at \((2,3)\), so \(x=2\).

    Algebra: add \(x\): \(3x-1=5\); add 1: \(3x=6\); divide: \(x=2\).

    -112345-2-1123456y = 2x – 1y = -x + 5(2, 3)
    The lines meet at (2, 3), so the equation \(2x-1=-x+5\) has solution \(x=2\).

    Problem 13: Solve for the height

    Multiply both sides by 2: \(2A=bh\). Divide by \(b\): \(h=\frac{2A}{b}\).

    Substitute: \(h=\frac{2(36)}{9}=8\).

    b = 9h = 8A = 36
    A triangle with \(b=9\), \(h=8\) and \(A=36\).

    Problem 14: Solve for width

    Subtract \(2l\): \(P-2l=2w\). Divide by 2: \(w=\frac{P-2l}{2}\).

    Substitute: \(w=\frac{50-30}{2}=10\).

    Problem 15: Standard form to y

    Subtract \(2x\): \(3y=12-2x\). Divide every term by 3: \(y=4-\frac{2}{3}x\).

    When \(x=3\): \(y=4-2=2\). Check: \(2(3)+3(2)=12\).

    Problem 16: Temperature formula

    Multiply by \(\frac{9}{5}\): \(\frac{9}{5}C=F-32\). Add 32: \(F=\frac{9}{5}C+32\).

    Substitute \(C=35\): \(F=63+32=95\), so \(35^{\circ}\text{C}=95^{\circ}\text{F}\).

    Problem 17: Taxi fare

    Let \(m\) be the miles: \(3.50+2.25m=17\).

    Subtract 3.50: \(2.25m=13.50\). Divide: \(m=6\) miles.

    Check: \(3.50+2.25(6)=17\).

    Problem 18: Gym plans

    Let \(n\) be months: \(60+25n=40n\).

    Subtract \(25n\): \(60=15n\). Divide: \(n=4\) months.

    Cost: \(60+25(4)=160\) and \(40(4)=160\), so \(\$160\) each.

    Problem 19: Garden fence

    Let \(w\) be the width, so the length is \(2w+3\).

    Perimeter: \(2(w+2w+3)=54\), so \(6w+6=54\). Subtract 6 and divide: \(w=8\).

    Width 8 ft, length 19 ft. Check: \(2(8+19)=54\).

    19 ft8 ftP = 54 ft
    The solved garden.

    Problem 20: Consecutive odd integers

    Let the integers be \(n\), \(n+2\), \(n+4\).

    \(3n+6=87\), so \(3n=81\) and \(n=27\).

    27, 29 and 31. Check: \(27+29+31=87\).

    Problem 21: Trapezoid base

    Multiply by 2: \(2A=(b_1+b_2)h\). Divide by \(h\): \(\frac{2A}{h}=b_1+b_2\). Subtract \(b_2\): \(b_1=\frac{2A}{h}-b_2\).

    Substitute: \(b_1=\frac{168}{8}-9=21-9=12\).

    b1 = 12b2 = 9h = 8
    The solved trapezoid: bases 12 and 9, height 8, area 84.

    Problem 22: Shopping with fractions

    Let \(s\) be his savings: \(s-\frac{s}{3}-\frac{s}{4}=35\).

    Multiply by 12: \(12s-4s-3s=420\), so \(5s=420\) and \(s=\$84\).

    Check: \(84-28-21=35\).

    See every chapter on the Algebra 1 math page.

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