Free Math Worksheets

- Where does the graph of \(f\) cross the y-axis? - How can the slope help you locate the output \(1.5\) from the y-intercept? - Which operation undoes subtracting \(3\)?

1. The function \(f(x) = 1.5x - 3\) has y-intercept \(-3\) and slope \(1.5 = \frac{3}{2}\). From \((0, -3)\), moving \(3\) units right changes \(y\) by \(1.5 \cdot 3 = 4.5\), reaching the point \((3, 1.5)\). Thus, \(x = 3\). 2. Algebraically, add \(3\) to both sides: \(1.5x = 4.5\). 3. Divide by \(1.5\): \(x = 3\). The algebraic solution agrees with the slope-and-intercept reasoning.

\(x = 3\)

- 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, \(S = \{\frac{5}{3}\}\).

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

- 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.

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

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

- Expand the two products separately. - Check whether the quadratic terms have opposite coefficients. - Solve the remaining linear equation.

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 \(5x^2 - 20 - 5x^2 + 5x = 5x - 20\). The student is correct. 4. Solve \(5x - 20 = 10\): \(5x = 30\), so \(x = 6\).

The student 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 \(-2x+18=10\). 2. Subtract \(18\): \(-2x=-8\), so \(x=4\). 3. For the second equation, expand: \(4x^2+20x+25-(4x^2-9)=74\). 4. Simplify: \(20x+34=74\), so \(20x=40\) and \(x=2\).

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

- Simplify each monomial quotient first. - Follow the order of operations before isolating the variable. - Check each solution against the nonzero restriction.

1. a) Simplify to \(3x = 18\), so \(x = 6\). 2. b) Simplify to \(5y + 13 = 48\). Then \(5y = 35\), so \(y = 7\). 3. c) Simplify the quotient to \(3z\): \(22z - 3z = 38\). Thus, \(19z = 38\), so \(z = 2\). 4. Each solution satisfies its nonzero restriction.

a) \(x = 6\) b) \(y = 7\) c) \(z = 2\)

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

1. a) Divide coefficients and powers: \((45x^2) \div (9x) = 5x\). 2. b) Substitute \(x = 6\): \(5 \cdot 6 + 6k = 84\). 3. Solve: \(30 + 6k = 84\), so \(6k = 54\) and \(k = 9\).

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

- Simplify the polynomial quotient first. - Divide each term in the numerator by \(9y\). - Then isolate \(y\) in the resulting linear equation.

1. Simplify the quotient: \((18y^2 - 9y) \div (9y) = 2y - 1\). 2. Solve \(14 + 2y - 1 = 20\), so \(13 + 2y = 20\). 3. Then \(2y = 7\), so \(y = \frac{7}{2}\). 4. The solution satisfies \(y \ne 0\).

\(y = \frac{7}{2}\)

- Simplify the two quotients separately. - Distribute the subtraction across the second expression. - Check the result against the nonzero restriction.

1. Simplify the first quotient: \((15x^2 + 10x) \div (5x) = 3x + 2\). 2. Simplify the second quotient: \((12x^2 - 18x) \div (6x) = 2x - 3\). 3. Solve \((3x + 2) - (2x - 3) = 1\). This gives \(x + 5 = 1\), so \(x = -4\). 4. The solution satisfies \(x \ne 0\).

\(x = -4\)

- 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.

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

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.

1. For a), multiply every term by \(10\): \(2(4z - 7) - (2z + 1) = 5z\). 2. Distribute and combine like terms: \(8z - 14 - 2z - 1 = 5z\), so \(6z - 15 = 5z\). Subtract \(5z\) and add \(15\): \(z = 15\). 3. For b), multiply every term by \(12\): \(12x - 3(2x - 3) = 2(5x + 1)\). 4. Distribute: \(12x - 6x + 9 = 10x + 2\). Combine like terms: \(6x + 9 = 10x + 2\). 5. Subtract \(6x\) and subtract \(2\): \(7 = 4x\). Divide by \(4\): \(x = \frac{7}{4} = 1.75\).

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

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 \(4\): \(x = 10\).

\(x = 10\)

- 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 \(2\): \(x = 2\).

\(x = 2\)

- 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.

1. For 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 \(x = -\frac{1}{5}\). 2. For 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) \(x = 4\)

- Identify the factor repeated in both terms on the left. - Factoring first may be more efficient than expanding every product. - Simplify the expression inside the brackets before solving. - Substitute your solution into the original equation to check it.

1. Factor out \(x - 3\): \((x - 3)[4(x + 5) - (4x + 2)] = 36\). 2. Simplify the bracket: \(4x + 20 - 4x - 2 = 18\). 3. Solve \(18(x - 3) = 36\): \(x - 3 = 2\), so \(x = 5\). 4. Check: \(4 \cdot 2 \cdot 10 - 22 \cdot 2 = 80 - 44 = 36\).

\(x = 5\)

- 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.

1. For a), simplify the inner parentheses: \(5x - [3x - 2x + 10] = 2x + 8\). 2. Combine inside the brackets and distribute the subtraction: \(5x - (x + 10) = 2x + 8\), so \(4x - 10 = 2x + 8\). 3. Subtract \(2x\) and add \(10\): \(2x = 18\). Divide by \(2\): \(x = 9\). 4. For b), multiply every term by the least common denominator, \(12\): \(4(2x - 5) - 3(3x - 7) = 2\). 5. Distribute and combine like terms: \(8x - 20 - 9x + 21 = 2\), so \(-x + 1 = 2\). 6. Subtract \(1\): \(-x = 1\). Multiply by \(-1\): \(x = -1\).

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

- 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.

1. 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 \(6\). 2. In line 2, Noah made a sign error. Distributing \(-2\) gives \(-2(2x - 3) = -4x + 6\). 3. Correct the equation: \(3(x + 4) - 2(2x - 3) = 6\). 4. Distribute: \(3x + 12 - 4x + 6 = 6\). 5. Combine like terms: \(-x + 18 = 6\). 6. Subtract \(18\): \(-x = -12\). Multiply by \(-1\): \(x = 12\).

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 \(s\).

1. The van's travel time is \(\frac{s}{v}\), and the motorcycle's travel time is \(\frac{s}{v + 15}\). 2. Therefore, \(\Delta t = \frac{s}{v} - \frac{s}{v + 15}\). 3. If \(s\) is doubled, both travel times double, so their difference also doubles. 4. For part c), the speeds are \(45\,\text{mph}\) and \(60\,\text{mph}\), and \(20\) minutes is \(\frac{1}{3}\) hour. 5. Write \(\frac{s}{45} - \frac{s}{60} = \frac{1}{3}\). 6. Multiply by \(180\): \(4s - 3s = 60\). 7. Thus, \(s = 60\).

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.

- First compare the runners' step lengths. - Use the number of steps each runner takes in the same time to compare their speeds. - At the catch-up point, the chaser has run the head start plus the other runner's distance. - Equal travel times mean the distance ratio equals the speed ratio.

1. Let \(a\) be Marcus's step length and \(e\) be Eli's step length. Since \(3a = 4e\), \(a = \frac{4}{3}e\). 2. In the same amount of time, Marcus covers \(4a\) and Eli covers \(5e\). Their speed ratio is therefore \(\frac{4a}{5e}\). 3. Substitute \(a = \frac{4}{3}e\): \(\frac{4(\frac{4}{3}e)}{5e} = \frac{16}{15}\). Marcus runs \(\frac{16}{15}\) as fast as Eli. 4. Let \(x\) yards be the distance Eli runs before being caught. Marcus runs \(x + 20\) yards in the same time. 5. Set the distance ratio equal to the speed ratio: \(\frac{x + 20}{x} = \frac{16}{15}\). 6. Cross-multiply: \(15(x + 20) = 16x\). 7. Solve: \(15x + 300 = 16x\), so \(x = 300\).

Eli will run \(300\) yards before Marcus catches him.