Free Math Worksheets

- Recall the addition and subtraction properties of \(0\). - Consider when a subtraction equation has a value of \(0\) or \(1\). - Check each solution in the original equation.

1. a) Subtracting \(0\) leaves a number unchanged, so \(x = 0\). 2. b) A number minus itself is \(0\), so \(x = 567\). 3. c) Solve \(888 - x = 1\): \(x = 888 - 1 = 887\). 4. d) Adding \(0\) leaves a number unchanged, so \(x = 999\).

a) \(x = 0\) b) \(x = 567\) c) \(x = 887\) d) \(x = 999\)

- Write an equation for the statement. - Use addition to undo subtraction. - Decide whether the original number must be greater or less than \(420\).

1. Write the equation \(x - 275 = 420\). 2. Add \(275\) to both sides: \(x = 420 + 275 = 695\).

The number is \(695\).

- Describe what number is missing in each equation. - Use addition or subtraction to undo the operation involving \(x\). - Pay attention to whether \(x\) is the first number or the number being subtracted. - Check each solution by substitution.

1. \(x = 400 - 130 = 270\). 2. \(x = 600 + 250 = 850\). 3. \(x = 900 - 720 = 180\). 4. \(x = 800 - 480 = 320\). 5. \(x = 750 - 300 = 450\).

1) \(x = 270\) 2) \(x = 850\) 3) \(x = 180\) 4) \(x = 320\) 5) \(x = 450\)

- Identify the missing addend in each equation. - Use subtraction to find a missing addend. - Try the same idea with a smaller equation such as \(2 + x = 5\). - Check by substitution.

1. In a), \(x = 400 - 340 = 60\). 2. In b), \(x = 290 - 250 = 40\). 3. In c), \(x = 580 - 510 = 70\). 4. In d), \(x = 800 - 720 = 80\).

a) \(x = 60\) b) \(x = 40\) c) \(x = 70\) d) \(x = 80\)

- Write an equation with a variable. - Use subtraction to undo addition. - Check by adding \(384\) to your answer.

1. Write the equation \(x + 384 = 721\). 2. Subtract \(384\) from both sides: \(x = 721 - 384 = 337\).

The number is \(337\).

- Write an equation with a variable for the unknown number. - Use subtraction to undo adding \(350\). - Check your result in the original equation.

1. Write the equation \(x + 350 = 820\). 2. Subtract \(350\) from both sides: \(x = 820 - 350\). 3. Compute: \(820 - 350 = 470\). 4. Check: \(470 + 350 = 820\).

The number is \(470\).

- Write an equation with a variable for the unknown number. - Use subtraction to undo adding \(237\). - Subtract by place value: hundreds, then tens, then ones. - Check your answer.

1. Write the equation \(x + 237 = 500\). 2. Subtract \(237\): \(x = 500 - 237 = 263\). 3. Check: \(263 + 237 = 500\).

\(263\)

- Write an equation with the unknown as the starting number. - Use addition to undo subtracting \(418\). - Decide whether the starting number must be greater or less than \(418\). - Check your answer.

1. Write \(x - 418 = 252\). 2. Add \(418\) to both sides: \(x = 252 + 418 = 670\). 3. Check: \(670 - 418 = 252\).

\(670\)

- Use the inverse of addition to isolate the variable. - Perform the same operation on both sides. - Substitute your answer into the original equation to check it.

1. Subtract \(9\) from both sides: \(x = 25 - 9\). 2. Calculate: \(x = 16\). 3. Check by substitution: \(16 + 9 = 25\), which is true.

\(x = 16\). The check gives \(16 + 9 = 25\).

- Use the inverse operation to isolate the unknown. - Think about how you would solve the same equation using whole numbers. - Align decimal points carefully when subtracting.

1. For a), subtract \(23.65\) from both sides: \(\square=41.10-23.65=17.45\). 2. For b), the missing subtrahend is \(60.20-18.47=41.73\), so \(\square=41.73\).

a) \(17.45\) b) \(41.73\)

- Use the inverse operation to isolate \(x\). - Convert fractions and decimals to a common form when needed. - Use a common denominator when subtracting fractions.

1. For a), subtract \(\frac{3}{8}\): \(x=\frac{5}{6}-\frac{3}{8}=\frac{20}{24}-\frac{9}{24}=\frac{11}{24}\). 2. For b), \(x=1.4-0.85=0.55\). 3. For c), add \(\frac{1}{4}\): \(x=0.3+0.25=0.55=\frac{11}{20}\).

a) \(x=\frac{11}{24}\) b) \(x=0.55\) c) \(x=0.55\), or \(x=\frac{11}{20}\)

- Use inverse operations to isolate \(x\). - In part b), remember that the coefficient of \(x\) is \(-1\). - Convert between fractions and decimals when useful.

1. For a), subtract \(\frac{7}{9}\): \(x=\frac{1}{3}-\frac{7}{9}=\frac{3}{9}-\frac{7}{9}=-\frac{4}{9}\). 2. For b), add \(2.5\): \(-x=1.3\). Multiply both sides by \(-1\): \(x=-1.3\). 3. For c), subtract \(1.1\): \(x=\frac{3}{5}-1.1=0.6-1.1=-0.5\).

a) \(x=-\frac{4}{9}\) b) \(x=-1.3\) c) \(x=-0.5\), or \(x=-\frac{1}{2}\)

- Use inverse operations to isolate \(x\). - Apply the same operation to both sides. - Simplify double negative signs before solving.

1. For a), subtract \(14.5\): \(x=-5.5-14.5=-20\). 2. For b), add \(20\): \(-x=55\), so \(x=-55\). 3. For c), rewrite as \(x+12=-8\). Then \(x=-20\). 4. For d), add \(4.2\): \(x=2.8+4.2=7\).

a) \(x=-20\) b) \(x=-55\) c) \(x=-20\) d) \(x=7\)

- Use inverse operations to find each missing number. - Convert fractions to a common denominator when needed. - Check each result by substitution.

1. For a), \(\Box=-1.2-(-3.4)=2.2\). 2. For b), \(\Box=-\frac{1}{6}+\frac{2}{3}=\frac{1}{2}\). 3. For c), \(\Box=1.5-0.8=0.7\).

a) \(2.2\) b) \(\frac{1}{2}\), or \(0.5\) c) \(0.7\)

- Convert fractions and decimals to a common form. - Simplify each side before isolating \(x\). - Use the inverse operation.

1. For a), write \(-\frac{3}{4}=-0.75\). Then \(x-0.4=-0.75\), so \(x=-0.35\). 2. For b), the right side is \(-0.4\), and \(\frac{2}{5}=0.4\). Thus \(x-0.4=-0.4\), so \(x=0\).

a) \(x=-0.35\), or \(x=-\frac{7}{20}\) b) \(x=0\)

- Identify the operation that must be undone. - Subtracting a negative number is equivalent to adding a positive number. - Compare your result with Leonie’s value.

1. To undo subtracting \(-2.4\), add \(-2.4\) to \(-1.1\). 2. Calculate: \(-1.1+(-2.4)=-3.5\). 3. This matches Leonie’s answer, so she is correct.

The inverse calculation is \(-1.1+(-2.4)=-3.5\). Leonie is correct.

- Use inverse operations to isolate the missing number. - In parts b) and c), think about what must be subtracted from the first number to reach the result. - Write the fractions with a common denominator before subtracting.

1. For part a), rewrite the equation as \(x+4.2=-1.5\). Subtract \(4.2\): \(x=-1.5-4.2=-5.7\). 2. For part b), solve \(2.5-x=6.1\): \(x=2.5-6.1=-3.6\). 3. For part c), write \(-\frac{1}{4}\) as \(-\frac{2}{8}\). Then \(x=-\frac{7}{8}-(-\frac{2}{8})=-\frac{5}{8}\).

a) \(-5.7\) b) \(-3.6\) c) \(-\frac{5}{8}\)

- Write an expression for the number that is \(2\) less than \(n\). - Combine the two subtractions. - Undo subtraction by adding the same amount to both sides.

1. Translate the statement into an equation: \((n-2)-10=88\). 2. Combine the subtractions: \(n-12=88\). 3. Add \(12\) to both sides: \(n=100\). 4. The whole number immediately after \(100\) is \(101\).

\(n=100\); the number immediately after it is \(101\).

- Use the inverse operation to isolate \(x\). - For an equation of the form \(a - x = b\), think about the difference between \(a\) and \(b\). - To subtract \(199\) mentally, you can subtract \(200\) and then add \(1\).

1. Subtract \(340\) from both sides: \(x = 1000 - 340 = 660\). 2. The missing subtrahend is the difference between \(1500\) and \(850\): \(x = 1500 - 850 = 650\). 3. Subtract \(199\) from both sides: \(x = 450 - 199 = 251\). 4. The missing subtrahend is the difference between \(2025\) and \(1975\): \(x = 2025 - 1975 = 50\).

a) \(x = 660\) b) \(x = 650\) c) \(x = 251\) d) \(x = 50\)

- Translate each sentence into an equation with \(x\). - Use the inverse operation to isolate \(x\). - Check your answer by substituting it into the original statement.

1. Write \(x + 156 = 400\). Subtract \(156\) from both sides: \(x = 400 - 156 = 244\). 2. Write \(x - 85 = 215\). Add \(85\) to both sides: \(x = 215 + 85 = 300\). 3. Write \(1000 - x = 634\). The missing subtrahend is \(x = 1000 - 634 = 366\).

a) \(x = 244\) b) \(x = 300\) c) \(x = 366\)

- Use the inverse operation to isolate \(x\). - Check each solution by substituting it into the original equation.

1. a) Subtract \(234\) from both sides: \(x = 500 - 234 = 266\). 2. b) Since \(782 - x = 400\), subtract \(400\) from \(782\): \(x = 382\). 3. c) Subtract \(56\) from both sides: \(x = 120 - 56 = 64\). 4. d) Add \(89\) to both sides: \(x = 111 + 89 = 200\).

a) \(x = 266\) b) \(x = 382\) c) \(x = 64\) d) \(x = 200\)

- Use the inverse operation for each equation. - Pay attention to whether the unknown is an addend, the minuend, or the subtrahend. - Check each solution by substitution.

1. a) Subtract \(1250\) from \(3100\): \(x = 3100 - 1250 = 1850\). 2. b) Add \(875\) to \(1425\): \(\square = 1425 + 875 = 2300\). 3. c) Subtract the difference from the minuend: \(x = 4005 - 2347 = 1658\). 4. Substituting each value into its original equation confirms all three solutions.

a) \(x = 1850\) b) \(\square = 2300\) c) \(x = 1658\)

- Use the inverse operation to isolate \(x\). - Ask which number must be added to the difference to recover the minuend. - Substitute your value for \(x\) to check the equation.

1. Add \(14{,}550\) to both sides: \(x = 5450 + 14{,}550\). 2. Calculate the sum: \(x = 20{,}000\).

\(x = 20{,}000\)

- Use the inverse of addition in part a. - In part b, the unknown is the subtrahend. - Check each value in the original equation.

1. a) Subtract \(857\): \(x = 2000 - 857 = 1143\). 2. b) Subtract the difference from the minuend: \(\square = 1500 - 634 = 866\). 3. Substitution confirms both solutions.

a) \(x = 1143\) b) \(\square = 866\)

- Find the missing addend in each equation. - Subtract the known addend from the sum. - Compare the two solutions. - Check by substitution.

1. In a), \(x = 520 - 460 = 60\). 2. In b), \(x = 450 - 390 = 60\). 3. Both equations have the same solution.

a) \(x = 60\) b) \(x = 60\) The solutions are equal.

- Identify whether \(x\) is the starting number or the number being subtracted. - If a number is subtracted from \(x\), undo the subtraction with addition. - If \(x\) is the number being subtracted, find what must be removed from the starting number. - Substitute each answer to check it.

1. a) Add \(120\): \(x = 250 + 120 = 370\). 2. b) Find the missing number being subtracted: \(x = 480 - 130 = 350\). 3. c) Add \(340\): \(x = 410 + 340 = 750\). 4. d) Find the missing number being subtracted: \(x = 860 - 590 = 270\).

a) \(x = 370\) b) \(x = 350\) c) \(x = 750\) d) \(x = 270\)

- Write each statement as an equation. - Use the inverse operation to isolate the unknown. - Check each answer in its equation.

1. a) Write \(x + 150 = 420\). Subtract \(150\): \(x = 420 - 150 = 270\). 2. b) Write \(x - 80 = 230\). Add \(80\): \(x = 230 + 80 = 310\). 3. c) The equation is \(610 + x = 900\). Subtract \(610\): \(x = 900 - 610 = 290\).

a) \(270\) b) \(310\) c) \(x = 290\)

- Use the inverse operation to isolate \(x\). - For \(a - x = b\), find the difference between \(a\) and \(b\). - Substitute each value to check the original equation.

1. Subtract \(450\) from both sides: \(x = 1000 - 450 = 550\). 2. The missing subtrahend is \(x = 820 - 540 = 280\). 3. Add \(280\) to both sides: \(x = 720 + 280 = 1000\). 4. Subtract \(1500\) from both sides: \(x = 2350 - 1500 = 850\).

a) \(x = 550\) b) \(x = 280\) c) \(x = 1000\) d) \(x = 850\)

- Use a variable for the unknown number. - Translate “subtracting \(14.2\) from a number” carefully. - Use the inverse operation to isolate the variable. - Pay attention to the signs of the numbers.

1. Let \(x\) be the unknown number. Write \(x - 14.2 = -5.8\). 2. Add \(14.2\) to both sides: \(x = -5.8 + 14.2\). 3. Calculate: \(x = 8.4\).

The number is \(8.4\).

- Use the inverse operation to isolate \(x\). - Check each result by substituting it into the original equation. - When several known numbers appear on one side, combine them first.

1. Add \(12\) to both sides: \(x=-5+12=7\). 2. Subtract \(3.4\) from both sides: \(x=-1.2-3.4=-4.6\). 3. Add \(\frac{2}{5}\) to both sides: \(x=\frac{1}{2}+\frac{2}{5}=\frac{5}{10}+\frac{4}{10}=\frac{9}{10}\). 4. First combine the known terms: \(7.5+(-10)=-2.5\). Then \(-2.5+x=0\), so \(x=2.5\).

1) \(x=7\) 2) \(x=-4.6\) 3) \(x=\frac{9}{10}\) 4) \(x=2.5\)

- Use an inverse operation to isolate \(x\). - Apply the same operation to both sides. - Track signs carefully when working with negative numbers. - Substitute each solution into its original equation to check it.

1. Subtract \(11\) from both sides: \(x = 4 - 11 = -7\). 2. Add \(6\) to both sides: \(x = -9 + 6 = -3\). 3. Add \(7\) to both sides: \(x = -2 + 7 = 5\). 4. Subtract \(20\) from both sides: \(-6 = x\), so \(x = -6\). 5. Rewrite the equation as \(x - 5 = -1\). Add \(5\) to both sides: \(x = 4\).

1) \(x = -7\) 2) \(x = -3\) 3) \(x = 5\) 4) \(x = -6\) 5) \(x = 4\)

- Would it help to rewrite every fraction in an equation with the same denominator? - When the denominators on both sides match, what must be true about the numerators? - Treat the resulting numerator equation as an equation with an unknown.

1. For a), rewrite \(\frac{2}{3}\) as \(\frac{12}{18}\). Then \(x+5=12\), so \(x=7\). 2. For b), rewrite \(\frac{1}{6}\) as \(\frac{5}{30}\). Then \(17-x=5\), so \(x=12\). 3. For c), rewrite \(1\) as \(\frac{42}{42}\). Then \(25+x-11=42\), so \(14+x=42\) and \(x=28\).

a) \(x = 7\) b) \(x = 12\) c) \(x = 28\)

- Use the inverse operation to isolate the missing value. - Convert the values in each equation to a common representation. - Check each answer by substitution.

1. For a), \(\frac{1}{2} = 0.5\). Subtract \(0.8\): \(0.5 - 0.8 = -0.3\). 2. For b), add \(\frac{3}{4}\) to both sides. Since \(\frac{3}{4} = 0.75\), the missing value is \(-0.5 + 0.75 = 0.25\). 3. For c), \(\frac{5}{8} = 0.625\). Subtract \(0.125\): \(0.625 - 0.125 = 0.5\).

a) \(-0.3\) b) \(0.25\) c) \(0.5\), or \(\frac{1}{2}\)

- Solve each equation separately. - Express all three solutions in the same form before comparing them. - Convert terminating decimals to fractions when useful.

1. Equation 1 gives \(x=0.75-0.5=0.25\). 2. Equation 2 gives \(x=\frac{5}{6}-\frac{7}{12}=\frac{10}{12}-\frac{7}{12}=\frac{1}{4}\). 3. Equation 3 gives \(x=0.125+\frac{1}{8}=0.125+0.125=0.25\). 4. Since \(\frac{1}{4}=0.25\), all three equations have the same solution.

Yes. All three equations have the solution \(x=0.25\), or \(x=\frac{1}{4}\).

- Rewrite the decimals as fractions first. - Make the denominators match within each equation. - Once the denominators match, solve the resulting equation for the numerator.

1. For a), rewrite \(0.3\) as \(\frac{3}{10}\) and \(\frac{4}{5}\) as \(\frac{8}{10}\). Then \(3+x=8\), so \(x=5\). 2. For b), rewrite \(0.5\) as \(\frac{1}{2}=\frac{8}{16}\) and \(\frac{7}{8}\) as \(\frac{14}{16}\). Then \(14-x=8\), so \(x=6\).

a) \(x=5\) b) \(x=6\)

- Rewrite the fractions with the same denominator. - Once the denominators match, what equation do the numerators satisfy? - Use the inverse operation to isolate \(x\).

1. Rewrite \(\frac{1}{4}\) as \(\frac{3}{12}\). 2. The equation becomes \(\frac{x}{12}-\frac{3}{12}=\frac{1}{12}\), so the numerator equation is \(x-3=1\). 3. Add \(3\) to both sides: \(x=4\).

\(x=4\)

- Simplify the double negative first. - Use the inverse operation to isolate \(x\). - Check the solution in the original equation.

1. Anna is correct because subtracting a negative number is equivalent to adding its opposite: \(x-(-15)=x+15\). 2. Ben is correct because subtracting \(15\) from both sides gives \(x=10-15\). 3. Therefore, \(x=-5\).

Both students are correct, and \(x=-5\).

- Use inverse operations to isolate \(x\). - What is a common denominator for \(4\) and \(6\)? - Rewrite the fractions before subtracting.

1. Add \(x\) to both sides and subtract \(\frac{1}{6}\) from both sides: \(x=\frac{3}{4}-\frac{1}{6}\). 2. Use denominator \(12\): \(\frac{3}{4}=\frac{9}{12}\) and \(\frac{1}{6}=\frac{2}{12}\). 3. Subtract: \(x=\frac{9}{12}-\frac{2}{12}=\frac{7}{12}\).

\(x=\frac{7}{12}\)

- First find the missing addend. - Use the value \(x=6\) to write a one-step equation. - Then substitute \(x=10\) into the completed expression.

1. Use the first input to find the missing number: \(2\times 6+\square=20\). 2. Then \(12+\square=20\), so \(\square=8\). 3. The expression is \(2x+8\). 4. For \(x=10\), \(2\times 10+8=28\).

\(28\)

- Use inverse operations to isolate \(x\). - Rewrite fractions with common denominators before adding or subtracting. - Simplify the final value of \(x\) when possible.

1. For a), isolate the amount being subtracted: \(x=\frac{8}{3}-\frac{5}{4}\). Using denominator \(12\), \(x=\frac{32}{12}-\frac{15}{12}=\frac{17}{12}\). 2. For b), add \(\frac{1}{2}\) to both sides: \(x=\frac{7}{10}+\frac{1}{2}=\frac{7}{10}+\frac{5}{10}=\frac{6}{5}\).

a) \(x=\frac{17}{12}\) b) \(x=\frac{6}{5}\)

- Can adding a whole number to \(345\) produce a smaller result? - Which operation undoes subtracting \(345\)? - Check each conclusion in the original equation.

1. For the first equation, subtracting \(345\) gives \(x = 210 - 345 = -135\). Since \(-135\) is not a whole number, the equation has no whole-number solution. 2. For the second equation, add \(345\) to both sides: \(x = 210 + 345 = 555\).

The equation \(x + 345 = 210\) has no solution in the whole numbers because its only solution is \(x = -135\). The solution of \(x - 345 = 210\) is \(x = 555\).

- Think about starting amounts that leave the same remainder after different amounts are taken away. - Use addition to undo subtraction.

1. Both equations have a difference of \(400\). Equation B subtracts a greater amount, so it must start with a greater value of \(x\). 2. For Equation A, add \(150\) to \(400\): \(x = 400 + 150 = 550\). 3. For Equation B, add \(250\) to \(400\): \(x = 400 + 250 = 650\). 4. Since \(650 > 550\), Equation B has the greater value of \(x\).

1. Equation B must have the greater value of \(x\) because it subtracts more but has the same difference. 2. Equation A: \(x = 550\); Equation B: \(x = 650\).

- Solve every equation for \(x\). - Record the four solutions. - Identify the solution that occurs only once.

1. A: \(x = 600 - 340 = 260\). 2. B: \(x = 850 - 590 = 260\). 3. C: \(x = 150 + 120 = 270\). 4. D: \(x = 730 - 470 = 260\). 5. Equation C does not belong because its solution is \(270\), while the others have solution \(260\).

Equation C does not belong. It has \(x = 270\); the other equations have \(x = 260\).

- Begin with the equation \(m - x = d\). - Test the relationship with a small example such as \(10 - x = 7\). - Identify which subtraction gives the amount that was removed.

1. From \(m - x = d\), subtract the difference from the starting value: \(x = m - d\). 2. To find the amount that was subtracted, subtract the difference from the starting value. 3. Substitute the values: \(x = 502 - 168 = 334\).

1. \(x = m - d\) 2. Subtract the difference from the starting value. 3. \(x = 334\)

- Evaluate the expression in parentheses first. - When a sum and one addend are known, subtract to find the other addend. - Consider what happens when the sum increases but \(85\) stays fixed.

1. \(130 - 45 = 85\), so the simplified equation is \(x + 85 = 210\). 2. Subtract \(85\): \(x = 210 - 85 = 125\). 3. Since \(x\) equals the sum minus the fixed addend \(85\), increasing the sum increases \(x\) by the same amount.

1. \(x + 85 = 210\) 2. \(x = 125\) 3. \(x\) would increase by the same amount as the right side.

- Use the inverse operation to isolate each variable. - Use a common denominator when adding or subtracting fractions. - Track negative signs carefully. - Convert a mixed number to an improper fraction before subtracting.

1. For a), subtract \(8.5\): \(x = 3.2 - 8.5 = -5.3\). 2. For b), add \(\frac{5}{6}\): \(y = \frac{1}{3} + \frac{5}{6} = \frac{2}{6} + \frac{5}{6} = \frac{7}{6} = 1\frac{1}{6}\). 3. For c), subtract \(12\): \(-z = 3.5\). Multiply by \(-1\): \(z = -3.5\). 4. For d), rewrite \(2\frac{1}{4}\) as \(\frac{9}{4}\). Then \(a = \frac{3}{8} - \frac{9}{4} = \frac{3}{8} - \frac{18}{8} = -\frac{15}{8} = -1\frac{7}{8}\).

a) \(x = -5.3\) b) \(y = \frac{7}{6} = 1\frac{1}{6}\) c) \(z = -3.5\) d) \(a = -\frac{15}{8} = -1\frac{7}{8}\)

- First evaluate the expression after “equals.” - Write an equation using \(x\) for the original number. - Use the inverse of addition.

1. Evaluate the stated result: \(150 - 38 = 112\). 2. Let the original number be \(x\). Then \(x + 45 = 112\). 3. Subtract \(45\): \(x = 112 - 45 = 67\).

\(67\)

- Begin with “starting value minus amount subtracted equals difference.” - Which operation undoes subtraction? - Substitute \(s = 38\) into your expression.

1. The relationship is \(\text{starting value} - s = 72\). 2. Therefore, the starting value is \(72 + s\). 3. When \(s = 38\), the starting value is \(72 + 38 = 110\). 4. In general, add the amount subtracted to the difference.

a) \(72 + s\) b) \(110\) c) Add the amount subtracted to the difference.