- Try a value for \(x\) and see whether the left side becomes \(17\). - Think about keeping the two sides of an equation equal as you work.

1. Subtract \(9\) from both sides: \(x+9-9=17-9\). 2. Simplify: \(x=8\). Substitution gives \(8+9=17\).

Subtract \(9\) from both sides; \(x=8\).

- What operation is being applied to \(x\)? - Use the inverse operation on both sides of the equation. - Substitute your result back into the original equation to check it.

1. Divide both sides by \(5\): \(x=7\). 2. Check: \(5\cdot7=35\).

\(x=7\)

- The right side is already a number; the left side needs a calculation. - Keep the proposed value of \(x\) unchanged during the check.

1. Replace \(x\) with \(5\): \(3\cdot5-4=15-4=11\). 2. Both sides equal \(11\), so \(x=5\) is a solution.

Yes. \(3\cdot5-4=11\).

- Which terms contain the same variable to the same power? - The constant term does not combine with an \(x\)-term. - Try the same input in both versions to see whether they agree.

1. The terms \(5x\) and \(-2x\) are like terms; their sum is \(3x\). 2. The equivalent equation is \(3x+7=16\).

\(3x+7=16\)

- Compare the variable terms on the two sides. Which side would be simpler to work with? - A change is valid only if both sides are treated equally.

1. One choice is to subtract \(x\) from both sides: \(4x+6-x=x+18-x\). 2. This gives \(3x+6=18\). Applying the same operation to both sides preserves the solutions. Subtracting \(4x\) from both sides is another valid choice.

\(3x+6=18\); both sides have been changed by the same amount.

- A given solution must make the equation true when substituted. - Simplify the quotient before substitution. - After substituting \(x = 6\), solve the resulting equation for \(k\).

a) Divide coefficients and powers: \((45x^2) \div (9x) = 5x\). b) Substitute \(x = 6\): \(5 \cdot 6 + 6k = 84\). Then \(30 + 6k = 84\), so \(6k = 54\) and \(k = 9\).

a) \(5x\) b) \(k = 9\)

- Find a common denominator for each equation. - Multiply every term on both sides by that denominator. - Combine the terms containing \(x\) after clearing the fractions.

1. For a), multiply every term by \(20\): \(8x + 5x = 260\). 2. Combine like terms: \(13x = 260\). Divide by \(13\): \(x = 20\). 3. For b), multiply every term by \(6\): \(7x - 3x = 48\). 4. Combine like terms: \(4x = 48\). Divide by \(4\): \(x = 12\).

a) \(x = 20\) b) \(x = 12\)

- Apply the distributive property to the entire quantity \(x+20\). - After distributing, compare the coefficients of \(x\) on the two sides. - Use the definition of \(x\) when interpreting the solution.

1. Distribute: \(15x+300=16x\). 2. Subtract \(15x\) from both sides: \(300=x\). 3. Therefore, \(x=300\). Since \(x\) represents Eli's distance before he is caught, Eli runs \(300\,\text{yd}\).

a) \(15x+300=16x\) b) \(x=300\) c) Eli runs \(300\,\text{yd}\) before Marcus catches him.

- Replace the predicted response with the target value. - Undo the intercept before undoing the rate. - Check the solution by substituting it back into the model.

1. Set the prediction equal to \(93\): \(93=7.5x+18\). 2. Subtracting \(18\) gives \(75=7.5x\). 3. Dividing by \(7.5\) gives \(x=10\).

\(10\) promotional posts

- Distribute before combining terms on the left side. - After simplifying, collect the variable terms on one side. - Use substitution in the original equation for the requested check.

1. Distribute: \(4x-12+7=2x+9\), so \(4x-5=2x+9\). 2. Subtract \(2x\): \(2x-5=9\). 3. Add \(5\): \(2x=14\), so \(x=7\). 4. Check: \(4(7-3)+7=23\) and \(2\cdot7+9=23\).

\(x=7\)

- How many terms are added on the left of the original equation? - What happens to each term when the whole side is divided by the same number? - Check the final value in the equation Jules started with.

1. Jules divided \(4x\) and \(20\) by \(4\) but did not divide \(8\) by \(4\). 2. Dividing each term gives \(x+2=5\), so \(x=3\). 3. Check in the original equation: \(4\cdot3+8=20\).

a) He forgot to divide \(8\) by \(4\); the correct equation is \(x+2=5\). b) \(x=3\); the original equation gives \(20=20\).

- Does Luis's new equation have the same solution as the original one? - Think about the entire equation, not only its left side. - A solution of one equivalent equation must also work in the other.

1. Multiplying only the left side by \(3\) changes the equation. Both sides must be multiplied by \(3\). 2. The equivalent equation is \(x+2=12\). 3. Subtract \(2\): \(x=10\). Checking gives \(\frac{10+2}{3}=4\).

Luis forgot to multiply the right side by \(3\). The solution is \(x=10\).

- Check your line formula against two plotted grid points. - Try comparing \(f(x)\) and \(g(x)\) at different inputs. Could their order change? - The meeting point does not need to lie inside the picture.

1. The line \(f\) rises \(1\) unit per unit to the right and meets the y-axis at \(1\), so \(f(x)=x+1\). 2. Equal outputs give \(x+1=2x-3\). Subtract \(x\) and add \(3\) to find \(x=4\). 3. Substitute \(x=4\) into either formula: \(y=5\). They meet at \((4,5)\), beyond the displayed graph.

a) Slope \(1\), y-intercept \(1\); \(f(x)=x+1\) b) \(x+1=2x-3\), so \(x=4\); intersection \((4,5)\)

- Use the least common denominator to clear the fractions in part a. - Apply the negative sign to the entire second numerator. - Expand the products in part b and look for terms that cancel.

a) Multiply by \(12\): \(3(3x - 1) - 4(x + 2) = -12\). Distribute and combine: \(9x - 3 - 4x - 8 = -12\), so \(5x - 11 = -12\). Then \(5x = -1\), so \(x = -\frac{1}{5}\). b) Expand: \(x^2 + 6x + 9 - (x^2 + 4x) = 17\). Combine like terms: \(2x + 9 = 17\). Then \(2x = 8\), so \(x = 4\).

a) \(x = -\frac{1}{5}\) b) \(x = 4\)

- Expand both products carefully. - Apply the subtraction sign to the entire second product. - Combine quadratic, linear, and constant terms separately. - Check whether the quadratic terms cancel.

1. Expand the square: \((x - 6)^2 = x^2 - 12x + 36\). 2. Expand the product: \((x + 2)(x - 2) = x^2 - 4\). 3. Substitute and distribute the subtraction: \(x^2 - 12x + 36 - x^2 + 4 = 20\). 4. Combine like terms: \(-12x + 40 = 20\). 5. Subtract \(40\): \(-12x = -20\). Divide by \(-12\): \(x = \frac{5}{3}\). 6. Therefore, \(S = \{\frac{5}{3}\}\).

\(S = \{\frac{5}{3}\}\)

- Distribute a subtraction sign to every term in the following parentheses. - Combine like terms before solving. - Check whether the quadratic terms in part b) cancel. - Group variable terms and constants separately.

1. For a), remove the parentheses and combine like terms: \(10x - 6 = 24\). Add \(6\): \(10x = 30\). Divide by \(10\): \(x = 3\). 2. For b), distribute the subtraction: \(2y^2 + 9y - 15 - 2y^2 + y - 5 = 60\). 3. The quadratic terms cancel, so \(10y - 20 = 60\). Add \(20\): \(10y = 80\). Divide by \(10\): \(y = 8\). 4. For c), distribute the subtraction and remove the parentheses: \(25 - 8z + 3 + 3z - 2 = 1\). 5. Combine like terms: \(-5z + 26 = 1\). Subtract \(26\): \(-5z = -25\). Divide by \(-5\): \(z = 5\).

a) \(x = 3\) b) \(y = 8\) c) \(z = 5\)

- Expand both sides completely before solving. - Look for identical terms that cancel from both sides. - Substitute the result into the original equation to check it.

a) The left side simplifies to \(4x^2 + 10x - 6\). The right side simplifies to \(4x^2 - 1 + 15 = 4x^2 + 14\). b) Solve \(4x^2 + 10x - 6 = 4x^2 + 14\). The quadratic terms cancel, giving \(10x = 20\), so \(x = 2\). c) Substitute \(x = 2\). The left side is \((8 - 2) \cdot 5 = 30\), and the right side is \(5 \cdot 3 + 15 = 30\). The solution checks.

a) \(4x^2 + 10x - 6 = 4x^2 + 14\) b) \(x = 2\) c) Both sides equal \(30\).

- Work from the innermost parentheses outward. - Multiply the entire equation by a common denominator to clear fractions. - Distribute subtraction signs carefully. - Check each result in the original equation.

a) Simplify the inner parentheses: \(5x - [3x - 2x + 10] = 2x + 8\). Combine inside the brackets and distribute the subtraction: \(5x - (x + 10) = 2x + 8\), so \(4x - 10 = 2x + 8\). Then \(2x = 18\), so \(x = 9\). b) Multiply every term by \(12\): \(4(2x - 5) - 3(3x - 7) = 2\). Distribute and combine: \(8x - 20 - 9x + 21 = 2\), so \(-x + 1 = 2\). Thus, \(x = -1\).

a) \(x = 9\) b) \(x = -1\)

- Expand the two products separately. - Compare the coefficients of the quadratic terms before simplifying. - Solve the equation that remains after like terms are combined.

1. Use the difference of squares: \(5(x - 2)(x + 2) = 5x^2 - 20\). 2. Expand the second product: \(-5x(x - 1) = -5x^2 + 5x\). 3. The left side becomes \(5x^2 - 20 - 5x^2 + 5x = 5x - 20\). Wiremu is correct. 4. Solve \(5x - 20 = 10\): \(5x = 30\), so \(x = 6\).

Wiremu is correct; the quadratic terms cancel. \(x = 6\).

- Expand each square or product before combining terms. - Distribute subtraction across every term in a polynomial. - Notice whether the quadratic terms cancel. - Substitute your solution into the original equation to check it.

1) Expand and simplify: \(2(x^2+6x+9)-2x^2-14x=10\), so \(-2x+18=10\). Then \(-2x=-8\), giving \(x=4\). 2) Expand: \(4x^2+20x+25-(4x^2-9)=74\). Simplify: \(20x+34=74\), so \(20x=40\) and \(x=2\).

1) \(x=4\) 2) \(x=2\)

- Put parentheses around each numerator before clearing the fractions. - Multiply every term, including the right side, by the common denominator. - Distribute the negative factor to both terms in the second numerator.

1. Multiply every term by the least common denominator, \(6\): \(3(x + 5) - 2(2x - 4) = 24\). 2. Distribute carefully: \(3x + 15 - 4x + 8 = 24\). 3. Combine like terms: \(-x + 23 = 24\). 4. Subtract \(23\): \(-x = 1\). Multiply by \(-1\): \(x = -1\). 5. The subtraction applies to the entire second numerator. Therefore, multiplying \(-(2x - 4)\) by \(2\) gives \(-4x + 8\), not \(-4x - 8\).

\(x = -1\). The subtraction sign must be distributed to every term in the second numerator.

- Multiply each equation by its least common denominator. - Apply the multiplication to every term on both sides. - Distribute any subtraction sign before combining like terms. - Isolate the variable after the fractions are cleared.

a) Multiply every term by \(4\): \(3x - 1 + 2(x + 2) = 20\). Distribute and combine like terms: \(5x + 3 = 20\). Then \(5x = 17\), so \(x = \frac{17}{5} = 3.4\). b) Multiply every term by \(6\): \(2(2y + 5) - (y - 1) = 12\). Distribute and combine like terms: \(4y + 10 - y + 1 = 12\), so \(3y + 11 = 12\). Then \(3y = 1\), so \(y = \frac{1}{3}\).

a) \(x = \frac{17}{5} = 3.4\) b) \(y = \frac{1}{3}\)

- Multiply every term by a common denominator to clear the fractions. - Include terms that do not originally have a denominator. - Distribute subtraction signs carefully. - Move variable terms to one side when they appear on both sides.

a) Multiply every term by \(10\): \(2(4z - 7) - (2z + 1) = 5z\). Distribute and combine: \(8z - 14 - 2z - 1 = 5z\), so \(6z - 15 = 5z\). Thus, \(z = 15\). b) Multiply every term by \(12\): \(12x - 3(2x - 3) = 2(5x + 1)\). Distribute: \(12x - 6x + 9 = 10x + 2\), so \(6x + 9 = 10x + 2\). Then \(7 = 4x\), giving \(x = \frac{7}{4} = 1.75\).

a) \(z = 15\) b) \(x = \frac{7}{4} = 1.75\)

- Find the least common denominator for each equation. - Multiply every term on both sides by that denominator. - Distribute carefully, especially after a subtraction sign. - Reduce each final fraction if possible.

1) Multiply by \(6\): \(2(2x - 5) + 3(x + 1) = 24\). Distribute and combine: \(4x - 10 + 3x + 3 = 24\), so \(7x - 7 = 24\). Thus, \(7x = 31\) and \(x = \frac{31}{7}\). 2) Multiply by \(12\): \(3(3z - 1) - 2(5 - z) = 24\). Distribute and combine: \(9z - 3 - 10 + 2z = 24\), so \(11z - 13 = 24\). Thus, \(11z = 37\) and \(z = \frac{37}{11}\).

1) \(x = \frac{31}{7}\) 2) \(z = \frac{37}{11}\)

- Multiply the entire equation by a common denominator. - Distribute the subtraction before the second fraction carefully. - Apply the multiplication to the right side as well. - Substitute your result into the original equation to check it.

1. Multiply every term by the least common denominator, \(6\): \(3(x + 4) - 2(x - 2) = 24\). 2. Distribute carefully: \(3x + 12 - 2x + 4 = 24\). 3. Combine like terms: \(x + 16 = 24\). 4. Subtract \(16\): \(x = 8\). 5. Check: \(\frac{8 + 4}{2} - \frac{8 - 2}{3} = 6 - 2 = 4\).

\(x = 8\)

- Use the least common denominator of \(3\) and \(4\). - Multiply the term \(x\) by the common denominator too. - Distribute the negative factor carefully. - Move variable terms to one side and constants to the other.

1. Multiply every term by the least common denominator, \(12\): \(12x - 4(2x + 1) = 3(x + 2)\). 2. Distribute: \(12x - 8x - 4 = 3x + 6\). 3. Combine like terms: \(4x - 4 = 3x + 6\). 4. Subtract \(3x\): \(x - 4 = 6\). 5. Add \(4\): \(x = 10\).

\(x = 10\)

- How does one full wheel turn relate to the distance traveled? - Which wheel makes more turns over the same distance? Why? - How can you express the difference in the two turn counts?

1. Let \(s\) be the course length in centimeters. The front wheel makes \(\frac{s}{20}\) turns, and the rear wheel makes \(\frac{s}{30}\) turns. 2. Their turn counts differ by \(15\), so \(\frac{s}{20}-\frac{s}{30}=15\). 3. Multiply by \(60\): \(3s-2s=900\), giving \(s=900\,\text{cm}\). 4. Convert to meters: \(900\,\text{cm}=9\,\text{m}\).

\(\frac{s}{20}-\frac{s}{30}=15\), so the course is \(9\,\text{m}\) long.

- Subtract the break from the total outing time. - Write a time expression for each direction. - The one-way distance is the same in both directions. - Clear the fractions before solving.

1. The group spends \(6 - 2 = 4\) hours hiking. 2. Let \(s\) miles be the one-way distance. 3. The uphill time is \(\frac{s}{3}\) hours, and the downhill time is \(\frac{s}{5}\) hours. 4. Write \(\frac{s}{3} + \frac{s}{5} = 4\). 5. Multiply by \(15\): \(5s + 3s = 60\). 6. Thus, \(8s = 60\), so \(s = 7.5\).

The one-way distance to the shelter is \(7.5\) miles.

- Write separate time expressions for the outbound and return trips. - Split the return distance into two equal parts. - Convert \(12\) minutes to hours. - Subtract the outbound time from the return time.

1. Let \(x\) miles be the one-way distance. 2. The outbound time is \(\frac{x}{15}\) hour. 3. The return time is \(\frac{x/2}{15} + \frac{x/2}{10} = \frac{x}{30} + \frac{x}{20}\) hours. 4. Convert the time difference: \(12\) minutes is \(\frac{1}{5}\) hour. 5. Write \(\frac{x}{30} + \frac{x}{20} - \frac{x}{15} = \frac{1}{5}\). 6. Multiply by \(60\): \(2x + 3x - 4x = 12\). 7. Therefore, \(x = 12\).

The one-way distance is \(12\) miles.

- Subtract the stop from the total trip time. - Find the boat's downstream and upstream speeds. - Write each travel time as distance divided by speed. - The one-way distance is the same in both directions.

1. The boat spends \(4 - 1 = 3\) hours traveling. 2. Its downstream speed is \(10 + 2 = 12\,\text{mph}\), and its upstream speed is \(10 - 2 = 8\,\text{mph}\). 3. Let \(d\) miles be the one-way distance. Write \(\frac{d}{12} + \frac{d}{8} = 3\). 4. Multiply by \(24\): \(2d + 3d = 72\). 5. Thus, \(5d = 72\), so \(d = 14.4\).

The landmark is \(14.4\) miles from the dock.

- Expand each product of binomials. - Pay close attention to the minus sign before the second product. - Notice what happens to the quadratic terms after combining like terms. - Isolate the variable after simplifying.

1. Expand the products: \((x + 3)(x - 4) = x^2 - x - 12\) and \((x - 5)(x + 2) = x^2 - 3x - 10\). 2. Substitute and distribute the subtraction: \(x^2 - x - 12 - (x^2 - 3x - 10) = 2\), so \(2x - 2 = 2\). 3. Add \(2\) to both sides: \(2x = 4\). 4. Divide by \(2\): \(x = 2\).

\(x = 2\)

- Simplify both sides of each equation before deciding what kind of solution set it has. - If the variable terms cancel, inspect the remaining statement instead of trying to divide by a variable coefficient. - Compare what a true numerical statement and a false numerical statement imply about the original equation.

a) Expand and simplify: \(4x-5=2x+7\). Subtract \(2x\) and add 5: \(2x=12\), so \(x=6\). This equation has one solution. b) Expand and simplify: \(2x+6=2x+7\). Subtract \(2x\): \(6=7\), which is false. This equation has no solution. c) Expand the left side: \(5x-3=5x-3\). Both sides are identical for every real \(x\), so the equation has infinitely many solutions.

a) \(x=6\); one solution b) No solution c) Infinitely many solutions

- Choose two points on the same line whose coordinates you can read exactly. - Check that a formula gives the right outputs at both points, not just at one. - Could the lines meet outside the part of the graph you can see?

1. The graph of \(f\) crosses the y-axis at \(1\) and rises \(2\) units for each \(1\) unit to the right, so \(f(x)=2x+1\). 2. The graph of \(g\) crosses the y-axis at \(7\) and falls \(1\) unit for each \(1\) unit to the right, so \(g(x)=-x+7\). 3. Set the formulas equal: \(2x+1=-x+7\). 4. Add \(x\) and subtract \(1\): \(3x=6\), so \(x=2\). 5. Substitute \(x=2\) into either formula to get \(y=5\). The intersection is \((2,5)\).

a) \(f\): y-intercept \(1\), slope \(2\); \(g\): y-intercept \(7\), slope \(-1\) b) \(f(x)=2x+1\) and \(g(x)=-x+7\) c) \(x=2\); intersection \((2,5)\)

- Focus on the sign of each product created by distributing \(-2\). - Rewrite the entire left side correctly before doing any balancing steps. - After simplifying both sides, collect the x-terms and constants separately.

1. The factor \(-2\) must multiply both terms inside the parentheses. In particular, \(-2\cdot(-4)=+8\), not \(-8\). 2. The correct expansion is \(5-6x+8=4x+9\), so \(13-6x=4x+9\). 3. Subtract \(9\) and add \(6x\): \(4=10x\). 4. Divide by \(10\): \(x=\frac{2}{5}\).

a) Jordan distributed the negative factor incorrectly: \(-2(-4)=+8\), not \(-8\). b) \(x=\frac{2}{5}\)

- Check whether the same operation was applied to every term on both sides. - Focus on the sign of each term when distributing \(-2\). - Rewrite the equation correctly before solving it.

a) In line 1, Noah multiplied the left side by the least common denominator, \(6\), but did not multiply the right side by \(6\). The right side should be \(6\). In line 2, he also made a sign error: \(-2(2x - 3) = -4x + 6\), not \(-4x - 6\). b) The corrected equation is \(3(x + 4) - 2(2x - 3) = 6\). Distribute: \(3x + 12 - 4x + 6 = 6\). Then \(-x + 18 = 6\), so \(-x = -12\) and \(x = 12\).

a) Noah failed to multiply the right side by \(6\), and he distributed \(-2\) incorrectly: \(-2(2x - 3) = -4x + 6\). b) \(x = 12\)

- Write the normal and delayed travel times separately. - Apply each speed to the correct percentage of the route. - Convert the delay to hours. - Set the delayed time minus the normal time equal to the delay.

1. Let \(d\) miles be the total route length. 2. The normal travel time is \(\frac{d}{60}\) hour. 3. The construction-day travel time is \(\frac{0.6d}{60} + \frac{0.4d}{40}\) hours. 4. Convert the delay: \(15\) minutes is \(0.25\) hour. 5. Write \(\frac{0.6d}{60} + \frac{0.4d}{40} - \frac{d}{60} = 0.25\). 6. Simplify: \(0.01d + 0.01d - \frac{d}{60} = 0.25\), so \(\frac{d}{300} = 0.25\). 7. Multiply by \(300\): \(d = 75\).

The train route is \(75\) miles long.

- Subtract the motorcycle's travel time from the van's travel time. - Think about how multiplying the same distance by \(2\) affects each travel time. - Convert \(20\) minutes to hours before writing the equation for part c). - Clear the denominators to solve for \(s\).

a) The van's travel time is \(\frac{s}{v}\), and the motorcycle's travel time is \(\frac{s}{v + 15}\). Therefore, \(\Delta t = \frac{s}{v} - \frac{s}{v + 15}\). b) If \(s\) is doubled while both speeds stay fixed, each travel time doubles. Their difference therefore doubles as well. c) The speeds are \(45\,\text{mph}\) and \(60\,\text{mph}\), and \(20\) minutes is \(\frac{1}{3}\) hour. Write \(\frac{s}{45} - \frac{s}{60} = \frac{1}{3}\). Multiplying by \(180\) gives \(4s - 3s = 60\), so \(s = 60\).

a) \(\Delta t = \frac{s}{v} - \frac{s}{v + 15}\) b) The time difference doubles. c) The route is \(60\) miles long.

- Write each cyclist's total travel time to the meeting point. - Include Alex's return distance and rest time. - Ben is still \(4\) miles short of the lake at the meeting. - Set the two total times equal.

1. Let \(s\) miles be the distance from Town A to Lake B. 2. Alex rides \(s\) miles to the lake and \(4\) miles back. His riding time is \(\frac{s + 4}{12}\) hours. 3. His \(10\)-minute rest is \(\frac{1}{6}\) hour, so his total time is \(\frac{s + 4}{12} + \frac{1}{6}\). 4. Ben rides \(s - 4\) miles before the meeting, so his time is \(\frac{s - 4}{8}\) hours. 5. The cyclists start together and meet at the same time, so \(\frac{s + 4}{12} + \frac{1}{6} = \frac{s - 4}{8}\). 6. Multiply by \(24\): \(2(s + 4) + 4 = 3(s - 4)\). 7. Solve: \(2s + 12 = 3s - 12\), so \(s = 24\).

Lake B is \(24\) miles from Town A.