Algebra Questions.

Explanation

Subtract 4 from both sides: 2x = 10. Divide by 2: x = 5.

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Add 2 to both sides: x/3 = 6. Multiply by 3: x = 18.

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Distribute: 5x − 10 = 3x + 8. Subtract 3x: 2x − 10 = 8. Add 10: 2x = 18. Divide by 2: x = 9.

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Both sides have the same denominator, so 3x − 1 = x + 5. Subtract x: 2x − 1 = 5. Add 1: 2x = 6. Divide: x = 3.

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Distribute: 3x + 6 − 5x + 5 = 2x + 7 → −2x + 11 = 2x + 7 → 4 = 4x → x = 1.

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Divide both sides by 4: x = 5.

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Subtract 7 from both sides: −3x = −18. Divide by −3: x = 6.

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For no solution, the coefficients of x must be equal but the constants different. Setting k = 3 gives 3x + 6 = 3x − 2, which simplifies to 6 = −2, a contradiction. So k = 3.

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Slope-intercept form: y = mx + b with m = 3 and b = −2 gives y = 3x − 2.

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Solve for y: −4y = −2x + 8 → y = (1/2)x − 2. The slope is 1/2.

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Slope = (13 − 5)/(6 − 2) = 2. Using y = 2x + b and point (2, 5): 5 = 4 + b → b = 1.

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Perpendicular lines have slopes that are negative reciprocals. The slope of y = 3x + 5 is 3, so the perpendicular slope is −1/3. Only y = −(1/3)x + 4 has this slope.

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3x + 2y = 6 → y = −(3/2)x + 3, so slope = −3/2. L: y − 4 = −(3/2)(x − 1) → y = −(3/2)x + 3/2 + 4 = −(3/2)x + 11/2.

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3 + 1.5m = 15 → 1.5m = 12 → m = 8 miles.

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Set y = 0: 0 = 2x − 6 → 2x = 6 → x = 3.

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The slope a = Δy/Δx = 10/4 = 5/2.

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Add the equations: 2x = 10 → x = 5.

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Add equations: 3x = 9 → x = 3. From x − y = 2: y = 1. So x + y = 4.

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From x − y = 2: x = y + 2. Substitute: 3(y + 2) + 2y = 16 → 5y + 6 = 16 → 5y = 10 → y = 2.

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The second equation is exactly 2 times the first, so they represent the same line. Every point on the line is a solution: infinitely many.

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For no solution, coefficient ratios must match but constants must not: a/3 = 2/1 → a = 6. Check: 6/3 = 2 = 2/1, but 6 ≠ 2 × 4 = 8. Confirmed no solution.

Explanation

Let a = apples, b = bananas. a + b = 20 and 0.5a + 0.3b = 8. Substituting b = 20 − a: 0.5a + 0.3(20 − a) = 8 → 0.2a = 2 → a = 10.

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Substitute y = 5: 2x + 1 = 5 → 2x = 4 → x = 2.

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x + y = 24 and x − y = 8. Adding: 2x = 32 → x = 16, y = 8. Product = 16 × 8 = 128.

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Add 6: 3x > 15. Divide by 3: x > 5.

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Subtract 4: −2x ≤ 6. Divide by −2 (flip the inequality): x ≥ −3.

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(65 + x)/2 ≥ 80 → 65 + x ≥ 160 → x ≥ 95.

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−6 < 2x − 4 < 6 → −2 < 2x < 10 → −1 < x < 5.

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4 − x > 2 → −x > −2 → x < 2. So x = 2 is NOT a solution (it makes the inequality false).

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"Less than or equal to 3" is written x ≤ 3.

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3x + 6 ≤ 2x + 10 → x ≤ 4.

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50n − 200 > 0 → n > 4. The smallest integer greater than 4 is 5.

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In y = mx + b, b is the y-intercept: the value when x = 0. Here, 7 is the cost before any hours are added, so it is the initial flat fee.

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Rate of change = (8 − 5)/(2 − 1) = 3. This holds for every consecutive pair in the table.

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2025 is 5 years after 2020. 12,000 + 5 × 450 = 12,000 + 2,250 = 14,250.

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Slope of m = −1/2. m: y = −(1/2)(x − 4). Set equal to ℓ: 2x = −x/2 + 2 → 5x/2 = 2 → x = 4/5, y = 8/5.

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−2x + 10 = 0 → −2x = −10 → x = 5.

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Fixed cost ($30) plus variable cost ($0.10 per text) gives C = 30 + 0.10t.

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Slope of f = (13 − 5)/(6 − 2) = 2. g(x) = 2x + 7 (since g(0) = 7). g(3) = 6 + 7 = 13.

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When x = 0 (no reading yet), y = 24. That is the total page count.

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Add 3 to both sides: 6x = 24. Divide by 6: x = 4.

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Distribute: 2x + 10 = 18 → 2x = 8 → x = 4.

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Slope-intercept form y = mx + b: m = −2 and b = 5 gives y = −2x + 5.

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y-intercept = 4, slope = 3. Slope-intercept form: y = 3x + 4.

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Distribute: 8x − 4 = 3x + 21 → 5x = 25 → x = 5.

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5x > 25 → x > 5.

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Slope = (0 − 6)/(3 − 0) = −6/3 = −2.

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Substitute x = 4: 4 + 2y = 10 → 2y = 6 → y = 3.

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Subtract the second equation from the first: 2x = 6 → x = 3.

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12h + 5 = 53 → 12h = 48 → h = 4.

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−4x + 20 = 0 → 4x = 20 → x = 5.

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2(3) + 3y = 18 → 6 + 3y = 18 → 3y = 12 → y = 4.

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For no solution the lines must be parallel (equal slope ratios, unequal constant ratio). Multiply the second equation by 2: 6x + 2y = 12. For the x-coefficients to match, k = 6. Since 8 ≠ 12, the lines are parallel and the system has no solution.

Explanation

3x + 6 ≥ 9 → x ≥ 1, or 3x + 6 ≤ −9 → 3x ≤ −15 → x ≤ −5. Solution: x ≤ −5 or x ≥ 1.

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25 + 0.15m = 40 → 0.15m = 15 → m = 100 miles.

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For infinitely many solutions the equation must be an identity: all coefficients and constants must match. So a = 3 (coefficient of x) and b = −7 (constant on the right equals −7 on the left).

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Slope of p = 2/3, so slope of q = −3/2. Using point-slope: y − 3 = −(3/2)(x − 4) → y = −(3/2)x + 6 + 3 = −(3/2)x + 9. The y-intercept is 9.

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n = 45/3 = 15.

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Rate of change = (5−2)/(1−0) = 3; y-intercept = 2. Equation: y = 3x + 2.

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−3x + 4 = 10 → −3x = 6 → x = −2.

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From the first equation: x = 1 + 3y. Substitute into the second: 2(1 + 3y) + y = 9 → 2 + 7y = 9 → 7y = 7 → y = 1.

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−2x + 7 = 3x − 8 → 15 = 5x → x = 3.

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5(3) + 2(−1) = 15 − 2 = 13.

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45(3) + 5n ≤ 200 → 135 + 5n ≤ 200 → 5n ≤ 65 → n ≤ 13. Maximum is 13 notebooks.

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6x − 12 − 2x − 2 = 4x − 14.

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Add the two equations: 3x = 12, so x = 4.

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Subtract 6: −3x > 6. Divide by −3 (flip the inequality): x < −2.

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Slope = (11 − 3) / (4 − 0) = 8 / 4 = 2.

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Set up: 3 + 2m = 19. Subtract 3: 2m = 16. Divide by 2: m = 8.

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Add 7 to both sides: 3x = 18. Divide by 3: x = 6.

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Distribute: 4x + 12 − 2x = 2x + 12.

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Let p = pens and n = notebooks. p + n = 8 and 2p + 4n = 22. From the first: n = 8 − p. Substitute: 2p + 4(8 − p) = 22 → 2p + 32 − 4p = 22 → −2p = −10 → p = 5.

Explanation

Rewrite in slope-intercept form: −2y = −3x + 12 → y = (3/2)x − 6. The slope is 3/2.

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Substitute x = 3: 4(3) + 2y = 20 → 12 + 2y = 20 → 2y = 8 → y = 4.

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Solve: 2x ≥ 8 → x ≥ 4. Among the choices, only x = 4 satisfies x ≥ 4.

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Let the integers be n, n+1, n+2. Then 3n + 3 = 48 → 3n = 45 → n = 15. The smallest is 15.

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f(7) = 3(7) − 4 = 21 − 4 = 17.

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Perpendicular lines have slopes that are negative reciprocals. The negative reciprocal of 2 is −1/2.

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Use y = mx + b: 5 = 3(2) + b → 5 = 6 + b → b = −1.

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From the second equation: y = 4x − 10. Substitute: 2x + 3(4x − 10) = 12 → 2x + 12x − 30 = 12 → 14x = 42 → x = 3. Then y = 4(3) − 10 = 2. So x + y = 3 + 2 = 5.

Explanation

Let the integers be n and n + 2. Then n(n + 2) = 48 → n² + 2n − 48 = 0 → (n + 8)(n − 6) = 0 → n = 6 (taking the positive). The integers are 6 and 8, and their sum is 14.

Explanation

Take the square root of both sides: x − 3 = ±5. So x = 3 + 5 = 8 or x = 3 − 5 = −2.

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Set up: 25 + 0.20m = 65 → 0.20m = 40 → m = 200.

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A system has infinitely many solutions only when the two equations describe the same line. Dividing the second equation by 2 gives 3x + 2y = 12, which must match 3x + ky = 12, so k = 2.

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x² = 4 → x = ±2. Both x = 2 and x = −2 are solutions.

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Slope a = (11 − 5)/(3 − 1) = 6/2 = 3. Using (1, 5): 5 = 3(1) + b → b = 2. So a + b = 3 + 2 = 5.

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FOIL: 2x·x + 2x·(−4) + 3·x + 3·(−4) = 2x² − 8x + 3x − 12 = 2x² − 5x − 12.

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Case 1: 2x − 6 = 10 → 2x = 16 → x = 8. Case 2: 2x − 6 = −10 → 2x = −4 → x = −2.

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Speed = 240/4 = 60 mph. Time = 390/60 = 6.5 hours.

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Let c be the weight of each carton. The total weight is represented by 4c + 7 = 39. Subtracting 7 gives 4c = 32, and dividing by 4 gives c = 8 pounds.

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Substitute −2 for x: f(−2) = 6 − 3(−2) = 6 + 6 = 12.

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If c is the number of additional crates, then 735 + 31c ≤ 1,200. Therefore, 31c ≤ 465, so c ≤ 15. The greatest possible whole number of crates is 15.

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Subtract (1/2)x from both sides to get (1/4)x + 5 = 12. Subtracting 5 gives (1/4)x = 7. Multiplying both sides by 4 gives x = 28.

Explanation

Let s be the number of 6-marker packs. The remaining 18 − s packs contain 4 markers each. Then 6s + 4(18 − s) = 92. Distributing gives 6s + 72 − 4s = 92, so 2s = 20 and s = 10.

Explanation

Write f(x) = mx + b. Since f(2) = 7, 2m + b = 7. Also, f(f(2)) = f(7) = 22, so 7m + b = 22. Subtracting the first equation from the second gives 5m = 15, so m = 3. Then 2(3) + b = 7, giving b = 1. Since f(0) = b, the answer is 1.