- 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) \(x = 80\)

- 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 \(16 + 9 = 25\).

- Which operation is being done to \(x\)? - What inverse operation would undo it on both sides?

1. Subtract \(7\) from both sides to get \(x=12\). 2. Check: \(12+7=19\).

\(x=12\); check: \(12+7=19\).

- Think about how to restore what was taken away. - Do the same operation to both sides of the equation.

1. Add \(5\) to both sides of \(y-5=9\). 2. The result is \(y=14\); checking gives \(14-5=9\).

\(y=14\). Adding \(5\) undoes subtracting \(5\).

- Look at the number attached to \(t\) by addition. - The same change to both sides keeps an equation balanced.

1. Subtract \(3\) from each side to keep them equal. 2. This gives \(t=11-3=8\).

Subtract \(3\) from each side. \(t=8\).

- Put the proposed number in place of the variable. - Decide whether the two sides now have the same value.

1. Substitute \(4\) for \(x\). 2. The left side becomes \(3+4=7\), equal to the right side, so it is a solution.

Yes. Substituting \(4\) gives \(3+4=7\).

- 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) \(41.73\)

- Identify whether \(x\) is being increased, decreased, or subtracted from a starting amount. - Use the inverse operation to isolate \(x\). - Substitute each solution into the original equation to check it.

1. For a), subtract \(14.5\): \(x=25.5-14.5=11\). 2. For b), subtract \(7\) from \(20\): \(x=20-7=13\). 3. For c), add \(12\): \(x=8+12=20\). 4. For d), subtract \(4.2\): \(x=7-4.2=2.8\).

a) \(x=11\) b) \(x=13\) c) \(x=20\) d) \(x=2.8\)

- Identify the operation performed on \(x\). - Use the inverse operation on both sides of the equation. - Check the proposed value in the original equation.

1. Anna is correct because adding \(15\) undoes subtracting \(15\). 2. Adding \(15\) to both sides gives \(x=10+15=25\). 3. Ben is also correct because the solution is \(x=25\).

Both students are correct, and \(x=25\).

- Identify the operation connected to the missing number. - Use the inverse operation to find each value. - Rewrite the fractions with a common denominator in part b).

1. For a), \(\square=5.6-3.4=2.2\). 2. For b), \(\square=\frac{1}{6}+\frac{2}{3}=\frac{1}{6}+\frac{4}{6}=\frac{5}{6}\). 3. For c), \(\square=1.5-0.8=0.7\).

a) \(2.2\) b) \(\frac{5}{6}\) c) \(0.7\)

- Convert fractions and decimals to the same form before calculating. - Simplify the numerical side of part b) first. - Use the inverse operation to isolate \(x\).

1. For a), write \(\frac{3}{4}=0.75\). Then \(x=0.75-0.4=0.35\). 2. For b), first simplify the right side: \(1.2-0.8=0.4\). Since \(\frac{2}{5}=0.4\), add \(0.4\) to both sides: \(x=0.8\).

a) \(x=0.35\), or \(x=\frac{7}{20}\) b) \(x=0.8\), or \(x=\frac{4}{5}\)

- Identify the operation performed on \(x\). - Write the inverse calculation using the number on the other side. - Substitute Mia’s value into the original equation to check it.

1. Adding \(2.4\) undoes subtracting \(2.4\). 2. Calculate: \(1.1+2.4=3.5\). 3. This matches Mia’s answer, so she is correct.

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

- Identify the role of the missing number in each subtraction equation. - Think about the relationship among the starting amount, the amount subtracted, and the result. - Rewrite the fractions with a common denominator in part c).

1. For part a), add \(4.2\): \(\square=1.5+4.2=5.7\). 2. For part b), find the difference between \(6.1\) and \(2.5\): \(\square=6.1-2.5=3.6\). 3. For part c), write \(\frac{1}{4}\) as \(\frac{2}{8}\). Then \(\square=\frac{7}{8}-\frac{2}{8}=\frac{5}{8}\).

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

- 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) \(x = 200\)

- Recall the addition and subtraction properties of \(0\). - For \(a-x=b\), find the difference between \(a\) and \(b\). - 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) \(x=999\)

- 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: \(x = 20{,}000\).

\(x = 20{,}000\)

- Represent each unknown with a variable before calculating. - Translate “sum” as addition and “difference between a number and \(20\)” as subtraction from the unknown. - Use the inverse addition or subtraction operation to isolate the variable. - Check each value in the equation you wrote.

1. For a), write \(x+30=75\). Subtract \(30\) from both sides to get \(x=45\). 2. For b), write \(y-20=10\). Add \(20\) to both sides to get \(y=30\). 3. Substitute each value into its equation to verify it.

a) Equation: \(x+30=75\); \(x=45\) b) Equation: \(y-20=10\); \(y=30\)

- Decide which quantity is the starting value and which quantity is being subtracted. - Use a variable for the unknown before solving. - Preserve the order of subtraction in each equation. - Check each value in the equation you wrote.

1. For a), write \(120-x=50\). The amount removed is \(120-50\), so \(x=70\). 2. For b), write \(y-45=100\). Add \(45\) to both sides to get \(y=145\). 3. Check each result in the original subtraction statement.

a) Equation: \(120-x=50\); \(x=70\) b) Equation: \(y-45=100\); \(y=145\)

- 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) \(\square = 866\)

- Represent the unknown number with a variable. - Write an equation that matches the statement. - Use addition to undo subtraction.

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

The number is \(695\).

- Simplify the side that contains no variable first. - Represent the unknown with a variable and write the equality. - Use the inverse of adding \(325\). - Check both sides after solving.

1. Evaluate the known side: \(900-150=750\). 2. Write \(x+325=750\). 3. Subtract \(325\) from both sides: \(x=425\). 4. Check that both sides equal \(750\).

Equation: \(x+325=750\) The number is \(425\).

- Identify whether \(x\) is an addend, a minuend, or a subtrahend. - Use the inverse operation to isolate \(x\). - 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) \(x=450\)

- Simplify the numerical expression first. - Replace that side with its value. - Then use an inverse operation to isolate \(x\). - Check each solution.

1. \(400+100=500\), so \(x-150=500\) and \(x=650\). 2. \(250+50=300\), so \(800-x=300\) and \(x=500\). 3. \(900-300=600\), so \(x+220=600\) and \(x=380\). 4. \(120+180=300\), so \(1000-x=300\) and \(x=700\).

1) \(x=650\) 2) \(x=500\) 3) \(x=380\) 4) \(x=700\)

- 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.

- Represent the unknown number with a variable. - Translate the statement into an addition equation. - Use subtraction to undo addition.

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

The number is \(337\).

- Simplify the fully numerical side first. - Write an equation with a variable for the unknown number. - Use the inverse of subtraction to isolate the variable. - Substitute the result to check.

1. Evaluate the known side: \(135+165=300\). 2. Write \(x-257=300\). 3. Add \(257\) to both sides: \(x=557\). 4. Check the equality in the original statement.

Equation: \(x-257=300\) The number is \(557\).

- Represent the unknown number with a variable. - Write an equation that matches the statement. - Use subtraction to undo adding \(350\).

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

The number is \(470\).

- Simplify the known sum before writing the equation. - Keep the variable in the subtracted position shown by the sentence. - Ask what amount must be removed from \(1000\) to reach the known difference. - Check the subtraction after solving.

1. Evaluate \(120+130=250\). 2. Write \(1000-x=250\). 3. The amount removed is \(1000-250\), so \(x=750\). 4. Check: \(1000-750=250\).

Equation: \(1000-x=250\) The number subtracted is \(750\).

- 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) \(x = 270\)

- Simplify the expression on the right first. - Write the resulting simpler equation. - Isolate \(x\) using subtraction. - Compare your result with \(400\).

1. Simplify the right side: \(900-140=760\). 2. Solve \(260+x=760\) by subtracting \(260\): \(x=760-260=500\). 3. Since \(500>400\), the missing number is greater than \(400\).

The missing number is \(500\). It is greater than \(400\).

- Simplify the side without the variable first. - Translate the remaining statement into an equation. - Use the inverse of subtracting \(360\). - Check the value in the equation.

1. Evaluate \(150+90=240\). 2. Write \(x-360=240\). 3. Add \(360\) to both sides: \(x=600\). 4. Check: \(600-360=240\).

Equation: \(x-360=240\) The number is \(600\).

- Evaluate the fully known expression for the difference first. - Represent the unknown number being subtracted with a variable. - Preserve the order of subtraction when you write the equation. - Check by substituting the result.

1. The known difference is \(2\times110=220\). 2. Write \(720-x=220\). 3. Solve for the subtracted number: \(x=720-220=500\). 4. Check: \(720-500=220\).

Equation: \(720-x=220\) The number subtracted is \(500\).

- Represent each unknown number with a variable. - Translate each verbal statement into an equation. - Use an inverse operation to isolate the unknown.

1. For a), \(x+150=420\), so \(x=420-150=270\). 2. For b), \(x-80=230\), so \(x=230+80=310\). 3. For c), \(610+x=900\), so \(x=900-610=290\).

a) \(x+150=420\); \(x=270\) b) \(x-80=230\); \(x=310\) c) \(610+x=900\); \(x=290\)

- Represent the unknown number with a variable. - Translate the question into an addition equation. - Use subtraction to undo adding \(237\).

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

\(263\)

- Represent the starting number with a variable. - Translate the question into a subtraction equation. - Use addition to undo subtracting \(418\).

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

\(670\)

- 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) \(x = 850\)

- Represent the unknown number with a variable. - Which number is being subtracted from which? - Use addition to undo subtraction.

1. Let \(x\) be the number. Write \(x-5.8=2.6\). 2. Add \(5.8\) to both sides: \(x=2.6+5.8=8.4\).

\(x-5.8=2.6\); the number is \(8.4\).

- Identify the operation performed on the variable. - Use the inverse operation on both sides. - Substitute each solution into its original equation to check it.

1. Subtract \(11\) from both sides: \(x=18-11=7\). 2. Subtract \(6\) from both sides: \(x=9-6=3\). 3. Add \(7\) to both sides: \(x=2+7=9\). 4. Subtract \(8\) from both sides: \(x=14-8=6\). 5. Subtract \(5\) from both sides: \(x=9-5=4\).

1) \(x=7\) 2) \(x=3\) 3) \(x=9\) 4) \(x=6\) 5) \(x=4\)

- Identify the amount already accounted for in the total distance. - Use the operation that undoes the addition in the equation.

1. Subtract \(2.75\) from both sides: \(x=8.50-2.75\). 2. Calculate: \(x=5.75\). 3. The distance is \(5.75\,\text{miles}\).

\(x=5.75\,\text{miles}\)

- Use the direction and label of the jump to connect the unknown start to the known landing. - Work backward from the landing value to find the start.

1. Let the unknown starting value be \(x\). 2. The diagram gives the equation \(x+6=19\). 3. Subtract \(6\): \(x=13\).

Equation: \(x+6=19\) Solution: \(x=13\)

- What amount must be removed from \(20\) to leave \(8\)? - Check the effect of the proposed addition on the variable term.

1. The equation says that subtracting \(z\) from \(20\) leaves \(8\). 2. The missing amount is \(20-8=12\), so \(z=12\). 3. Adding \(8\) to both sides gives \(28-z=16\), which still has \(z\) subtracted.

\(z=12\). Adding \(8\) to both sides does not remove the \(-z\) term.

- Read the signed amount written on the jump. - Work backward from the landing point to find the starting point.

1. The jump adds \(7\), so the starting number \(x\) satisfies \(x+7=18\). 2. Subtract \(7\) from both sides: \(x=11\).

Starting number: \(11\). Equation: \(x+7=18\).

- Start with the unknown number that was already on the shelf. - Which operation undoes the extra books being added?

1. Starting books plus new books gives \(b+6=22\). 2. Subtract \(6\) from both sides to find \(b=16\).

\(b+6=22\); \(b=16\) books.

- Try Amir's value in the original equation. - How far apart are \(15\) and \(9\) on a number line?

1. The equation says a number is removed from \(15\) to leave \(9\), not added to \(15\). 2. The removed number is \(15-9=6\). 3. Checking gives \(15-6=9\).

Amir added instead of finding the amount removed. \(x=6\).

- How do the two bottom bricks make the top brick in each wall? - Work out what the missing brick must contribute to the sum. - Compare the two bottom rows after finding both blanks.

1. In wall a), \(a+6=15\). Subtract \(6\) from both sides to get \(a=9\). 2. In wall b), \(b+9=15\). Subtract \(9\) from both sides to get \(b=6\). 3. The top stays \(15\). The known bottom brick increases by \(3\), so the missing bottom brick decreases by \(3\) to keep the sum unchanged.

a) \(a+6=15\), so \(a=9\). b) \(b+9=15\), so \(b=6\). The known bottom number rises by \(3\), so the missing one falls by \(3\) to keep the sum \(15\).

- In each panel, notice where the jump begins and ends. - How can you check a missing starting number using the landing number? - Compare the directions of the two jumps.

1. In a), \(a-5=9\), so \(a=14\). 2. In b), \(9+5=b\), so \(b=14\). 3. Subtracting \(5\) from \(14\) lands at \(9\), and adding \(5\) to \(9\) returns to \(14\). The two jumps undo each other.

a) \(14\); equation: \(a-5=9\). b) \(14\); equation: \(9+5=b\). The jumps are opposites: subtracting \(5\) and then adding \(5\) returns to the start.

- 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) \(x = 28\)

- Evaluate the powers before solving the equation. - Simplify both sides. - Use an inverse operation to isolate \(n\).

1. Evaluate the powers: \(2^5=32\) and \(5^2=25\). 2. Simplify the equation: \(n+32=137\). 3. Subtract \(32\) from both sides: \(n=105\).

\(n=105\)

- Use inverse addition or subtraction to isolate the unknown. - For part b), combine the known fractional terms before solving. - Use a common denominator for mixed-number calculations. - Check each value in its original equation.

1. For a), the amount subtracted is the difference between the starting value and the result: \(x=5\frac{1}{2}-2\frac{3}{4}=2\frac{3}{4}\). 2. For b), \(1\frac{1}{3}-\frac{1}{6}=\frac{8}{6}-\frac{1}{6}=\frac{7}{6}\). The equation becomes \(x+\frac{7}{6}=3\). 3. Subtract \(\frac{7}{6}\) from both sides: \(x=\frac{18}{6}-\frac{7}{6}=\frac{11}{6}=1\frac{5}{6}\).

a) \(x=2\frac{3}{4}\) b) \(x=1\frac{5}{6}\)

- 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 \(0.3\): \(0.5-0.3=0.2\). 2. For b), add \(\frac{1}{4}=0.25\): \(0.5+0.25=0.75\). 3. For c), \(\frac{5}{8}=0.625\). Subtract \(0.125\): \(0.625-0.125=0.5\).

a) \(0.2\) b) \(0.75\), or \(\frac{3}{4}\) c) \(0.5\), or \(\frac{1}{2}\)

- 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 \(x=\frac{11}{20}\)

- 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 \(x=\frac{1}{4}\).

- Identify the operation being performed on \(x\) in each equation. - Use the inverse operation to isolate \(x\). - Convert between fractions and decimals only when it makes the subtraction easier.

1. For a), subtract \(\frac{1}{3}\): \(x=\frac{7}{9}-\frac{1}{3}=\frac{7}{9}-\frac{3}{9}=\frac{4}{9}\). 2. For b), subtract \(1.2\) from \(2.5\): \(x=2.5-1.2=1.3\). 3. For c), subtract \(\frac{3}{5}\): \(x=1.1-\frac{3}{5}=1.1-0.6=0.5\).

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

- 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 fully known side before isolating the variable. - Use a common denominator in part a). - In part b), think about how much must be removed from the starting amount. - Check each solution in its original equation.

1. For a), \(\frac{2}{3}+\frac{1}{2}=\frac{7}{6}\). Add \(\frac{1}{6}\) to both sides: \(x=\frac{8}{6}=\frac{4}{3}\). 2. For b), \(\frac{1}{8}+\frac{1}{8}=0.25\). Hence \(0.75-x=0.25\) and \(x=0.75-0.25=0.5\).

a) \(x=\frac{4}{3}\), or \(1\frac{1}{3}\) b) \(x=0.5\), or \(\frac{1}{2}\)

- Simplify the known quantities before applying an inverse operation. - Convert decimals and fractions only if it helps the calculation. - Use a common denominator in part b). - Substitute each result to check its equation.

1. For a), \(\frac{1}{5}+0.1=0.3\), so the right side is \(0.8\). Then \(x+0.3=0.8\), giving \(x=0.5\). 2. For b), \(\frac{1}{4}-\frac{1}{12}=\frac{3}{12}-\frac{1}{12}=\frac{1}{6}\). The equation becomes \(\frac{2}{3}=x+\frac{1}{6}\). 3. Therefore, \(x=\frac{4}{6}-\frac{1}{6}=\frac{1}{2}\).

a) \(x=0.5\), or \(\frac{1}{2}\) b) \(x=\frac{1}{2}\)

- Combine the two known changes before solving for the variable. - After simplifying, identify the single operation being performed on \(x\). - Use the inverse operation on both sides, then check in the original sequence.

1. Adding \(8\) and then subtracting \(14\) produces an overall decrease of \(14-8=6\), so the puzzle is equivalent to \(x-6=10\). 2. Add \(6\) to both sides: \(x=16\). 3. Check in the original sequence: \(16+8-14=10\).

Equivalent one-step equation: \(x-6=10\) Solution: \(x=16\)

- 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) \(x=\frac{6}{5}\)

- Write an expression for the number that is \(2\) less than \(n\). - Combine the two subtractions before solving. - 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 \(101\).

- Think about the smallest possible value of \(x+345\) when \(x\) is a whole number. - Use addition to undo subtracting \(345\). - Check the second solution in its original equation.

1. If \(x\) is a whole number, then \(x\ge 0\), so \(x+345\ge345\). It cannot equal \(210\). Therefore, the first 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 adding a whole number to \(345\) cannot produce a value less than \(345\). The solution of \(x-345=210\) is \(x=555\).

- Combine the two known changes before solving. - After simplification, identify the single addition operation on the unknown. - Use the inverse operation on both sides. - Check the answer using the original two changes.

1. The net change is \(120-50=70\), so write \(x+70=600\). 2. Subtract \(70\) from both sides: \(x=530\). 3. Check: \(530+120-50=600\).

Equivalent one-step equation: \(x+70=600\) Starting number: \(530\)

- Translate each statement into its own equation before calculating. - Use the inverse of subtraction for Luke’s equation and the inverse of addition for Marie’s. - Keep the two variables separate until both equations are solved. - Compare the solutions and then find their difference.

1. Luke’s statement gives \(x-240=360\). Add \(240\) to both sides: \(x=600\). 2. Marie’s statement gives \(y+180=800\). Subtract \(180\) from both sides: \(y=620\). 3. Marie’s number is greater, and \(620-600=20\).

Luke: \(x-240=360\), so \(x=600\). Marie: \(y+180=800\), so \(y=620\). Marie’s number is greater by \(20\).

- Combine the subtraction and addition into one net change. - Write the simplified relation with a variable before solving. - Use the inverse of the remaining subtraction. - Verify the value with the original sequence.

1. The subtraction of \(245\) is partly offset by adding \(112\). The overall decrease is \(245-112=133\), so write \(x-133=500\). 2. Add \(133\) to both sides: \(x=633\). 3. Check: \(633-245+112=500\).

Equivalent one-step equation: \(x-133=500\) Starting number: \(633\)

- 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 \(x = 260\).

- Combine the two known additions before solving. - After simplification, identify the single operation on \(x\). - Use the inverse operation on both sides. - Check using the original two additions.

1. Combine the known changes: \(15+22=37\). 2. Write the equivalent one-step equation \(x+37=44\). 3. Subtract \(37\) from both sides: \(x=7\). 4. Check in the original sequence: \(7+15+22=44\).

Equivalent one-step equation: \(x+37=44\) \(x=7\)

- 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.

- Identify whether the variable is an addend, a minuend, or a subtrahend. - Use a common denominator when adding or subtracting fractions. - Convert the mixed number to an improper fraction in part d).

1. For a), subtract \(3.2\): \(x=8.5-3.2=5.3\). 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), find the missing subtrahend: \(z=15.5-12=3.5\). 4. For d), rewrite \(2\frac{1}{4}\) as \(\frac{18}{8}\). Then \(a=\frac{18}{8}-\frac{3}{8}=\frac{15}{8}=1\frac{7}{8}\).

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 \(45\): \(x = 112 - 45 = 67\).

\(67\)

- Use the inverse operation to isolate \(x\). - Rewrite fractions with a common denominator in part 3. - Combine the known numbers before solving part 4.

1. Subtract \(12\) from both sides: \(x=19-12=7\). 2. Subtract \(3.4\): \(x=8-3.4=4.6\). 3. Rewrite \(\frac{2}{5}\) as \(\frac{4}{10}\). Then \(x=\frac{9}{10}-\frac{4}{10}=\frac{1}{2}\). 4. First combine the known terms: \(7.5+2.5=10\). Then \(10+x=12.5\), so \(x=2.5\).

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

- 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.

- Write an equation for the two addends and their sum. - Which operation undoes addition? - Substitute \(s = 42\) into your expression.

1. If the unknown addend is \(x\), then \(15 + x = s\), so \(x = s - 15\). 2. When \(s = 42\), \(x = 42 - 15 = 27\). 3. Subtract the known addend from the sum.

1. \(s - 15\) 2. \(27\) 3. Subtract the known addend from the sum.

- Substitute each proposed value into the left side of the original equation. - Rewrite the fractions with denominator \(8\) before subtracting. - Compare each result with the right side.

1. For \(y=\frac{3}{4}\), \(\frac{3}{4}-\frac{3}{8}=\frac{6}{8}-\frac{3}{8}=\frac{3}{8}\), not \(\frac{5}{8}\). 2. For \(y=1\), \(1-\frac{3}{8}=\frac{8}{8}-\frac{3}{8}=\frac{5}{8}\). 3. Therefore, \(y=1\) solves the equation.

\(y=1\)