Math – Algebra

Elite Worked Examples (15) — (Hard to MIT Level)

Example 1 (Hard)

A system of equations is given by: \begin{cases} \frac{1}{2}(ax – By) = 8 \\ 3x – 2y = 20 \end{cases} If the system has infinitely many solutions, what is the value of \( a + B \)?

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1. Multiply the first equation by 2 to clear the fraction: \( ax – By = 16 \).
2. For infinitely many solutions, the ratio of coefficients and constants must be equal: \( \frac{a}{3} = \frac{-B}{-2} = \frac{16}{20} \).
3. Simplify the constant ratio: \( \frac{16}{20} = 0.8 \).
4. Solve for \(a\): \( \frac{a}{3} = 0.8 \Rightarrow a = 2.4 \).
5. Solve for \(B\): \( \frac{B}{2} = 0.8 \Rightarrow B = 1.6 \).
6. Sum: \( 2.4 + 1.6 = 4 \).

Answer: 4

Example 2 (Hard)

Line \( \ell \) in the xy-plane passes through the point \((3, 5)\) and is perpendicular to the line \( 2x – 5y = 10 \). If line \( \ell \) intersects the x-axis at \((k, 0)\), find \(k\).

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1. Find the slope of the given line: \( -5y = -2x + 10 \Rightarrow y = \frac{2}{5}x – 2 \). Slope is \( \frac{2}{5} \).
2. The perpendicular slope is the negative reciprocal: \( m = -\frac{5}{2} \).
3. Write the equation for line \( \ell \) using point-slope: \( y – 5 = -\frac{5}{2}(x – 3) \).
4. To find the x-intercept, set \( y = 0 \): \( -5 = -\frac{5}{2}(x – 3) \).
5. Multiply by \( -\frac{2}{5} \): \( 2 = x – 3 \Rightarrow x = 5 \).

Answer: 5

Example 3 (Hard)

Solve for all values of \(x\) that satisfy: \( |3x – 2| + 4 = 1 \).

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1. Isolate the absolute value: \( |3x – 2| = 1 – 4 \).
2. Simplify: \( |3x – 2| = -3 \).
3. Critical Concept: An absolute value represents a distance and can never result in a negative number.
4. Therefore, there is no value of \(x\) that makes this equation true.

Answer: No Solution

Example 4 (Hard)

A shipping service charges a basic fee of \( \$12.00 \) plus \( \$0.50 \) per ounce for the first 16 ounces. For every ounce over 16, the rate is \( \$0.35 \) per ounce. If the total charge was \( \$24.15 \), what was the weight of the package in ounces?

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1. Cost of the first 16 oz: \( 12.00 + 0.50(16) = 12 + 8 = \$20.00 \).
2. Subtract this from the total: \( 24.15 – 20.00 = \$4.15 \) remaining.
3. Find additional ounces: \( \frac{4.15}{0.35} \). Use long division or a calculator: \( 415 / 35 = 11.85… \) wait, let’s re-check the math.
4. \( 4.15 \div 0.35 \approx 11.85 \). Total weight = \( 16 + 11.85 \).

Answer: 27.85 ounces

Example 5 (Hard)

If \( \frac{2x}{y} = 4 \) and \( \frac{y}{3z} = 2 \), what is the value of \( \frac{x}{z} \)?

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1. Solve for \(x\) in the first eq: \( 2x = 4y \Rightarrow x = 2y \).
2. Solve for \(y\) in the second eq: \( y = 6z \).
3. Substitute \(y\) into the \(x\) equation: \( x = 2(6z) = 12z \).
4. Divide both sides by \(z\): \( \frac{x}{z} = 12 \).

Answer: 12

Example 6 (Ivy)

The function \( f(x) = mx + b \) satisfies \( f(f(x)) = 9x + 8 \). If \( m > 0 \), find the value of \( f(2) \).

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1. \( f(f(x)) = m(mx + b) + b = m^2x + mb + b \).
2. Compare this to \( 9x + 8 \): \( m^2 = 9 \) and \( mb + b = 8 \).
3. Since \( m > 0 \), \( m = 3 \).
4. Plug \(m\) into the second part: \( 3b + b = 8 \Rightarrow 4b = 8 \Rightarrow b = 2 \).
5. The function is \( f(x) = 3x + 2 \).
6. \( f(2) = 3(2) + 2 = 8 \).

Answer: 8

Example 7 (Ivy)

A region in the xy-plane is defined by the system of inequalities: \[ y \ge |x| \quad \text{and} \quad y \le 4 \] Find the area of the resulting geometric shape.

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1. The inequality \( y \ge |x| \) creates a V-shape starting at \((0,0)\).
2. The inequality \( y \le 4 \) caps this V-shape at a height of 4.
3. This forms a triangle. The height is 4 (from \(y=0\) to \(y=4\)).
4. Find the base: At \( y = 4 \), \( |x| = 4 \), so \( x = 4 \) and \( x = -4 \). The base length is \( 4 – (-4) = 8 \).
5. Area = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 4 = 16 \).

Answer: 16

Example 8 (Ivy)

Find the value of \(k\) such that the line \( y = kx \) is tangent to the parabola \( y = x^2 + 4 \).

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1. Set the equations equal: \( x^2 + 4 = kx \Rightarrow x^2 – kx + 4 = 0 \).
2. A line is tangent if there is exactly one solution (discriminant = 0).
3. \( D = b^2 – 4ac \Rightarrow (-k)^2 – 4(1)(4) = 0 \).
4. \( k^2 – 16 = 0 \Rightarrow k = \pm 4 \).

Answer: \(\pm 4\)

Example 9 (Ivy)

If \( a + b = 10 \) and \( a^2 – b^2 = 40 \), what is the value of \( 3a – 2b \)?

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1. Use the difference of squares: \( a^2 – b^2 = (a+b)(a-b) = 40 \).
2. Substitute \( a+b = 10 \): \( 10(a-b) = 40 \Rightarrow a – b = 4 \).
3. Solve the system: \( (a+b) + (a-b) = 14 \Rightarrow 2a = 14 \Rightarrow a=7 \).
4. Then \( b=3 \).
5. Calculate \( 3(7) – 2(3) = 21 – 6 = 15 \).

Answer: 15

Example 10 (Ivy)

For what value of \(c\) will the system of equations have no solution? \[ \begin{cases} cx – 4y = 10 \\ (c-3)x – 2y = 12 \end{cases} \]

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1. For no solution, the slopes must be equal: \( \frac{c}{c-3} = \frac{-4}{-2} \).
2. Simplify: \( \frac{c}{c-3} = 2 \).
3. Solve for \(c\): \( c = 2(c – 3) \Rightarrow c = 2c – 6 \Rightarrow c = 6 \).
4. Verify constants: If \(c=6\), eq 1 is \(6x-4y=10\) and eq 2 is \(3x-2y=12\).
5. Multiply eq 2 by 2: \(6x-4y=24\). Since \(10 \neq 24\), it’s a parallel mismatch (no solution).

Answer: 6

Example 11 (MIT)

A car travels from point A to B at 40 mph and returns from B to A at 60 mph. What is the average speed for the entire round trip?

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1. Let the distance between A and B be \( D \).
2. Time to go: \( t_1 = D/40 \). Time to return: \( t_2 = D/60 \).
3. Total distance = \( 2D \). Total time = \( \frac{D}{40} + \frac{D}{60} = \frac{3D+2D}{120} = \frac{5D}{120} = \frac{D}{24} \).
4. Average Speed = \( \frac{\text{Total Distance}}{\text{Total Time}} = \frac{2D}{D/24} = 2 \times 24 = 48 \).

Answer: 48 mph

Example 12 (MIT)

Two pipes can fill a tank together in 4 hours. If Pipe A fills the tank 6 hours faster than Pipe B alone, how many hours does it take Pipe A to fill the tank?

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1. Let \(x\) be the hours for Pipe A. Then Pipe B takes \(x+6\) hours.
2. Rates: \( \frac{1}{x} + \frac{1}{x+6} = \frac{1}{4} \).
3. Common denominator: \( \frac{x+6+x}{x(x+6)} = \frac{1}{4} \Rightarrow \frac{2x+6}{x^2+6x} = \frac{1}{4} \).
4. Cross-multiply: \( 8x + 24 = x^2 + 6x \Rightarrow x^2 – 2x – 24 = 0 \).
5. Factor: \( (x-6)(x+4) = 0 \). Since time must be positive, \( x=6 \).

Answer: 6 hours

Example 13 (MIT)

Find the coordinates of the point on the line \( y = 2x + 5 \) that is closest to the origin \((0,0)\).

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1. The closest point is found by a line perpendicular to \( y = 2x + 5 \) passing through \((0,0)\).
2. Perpendicular slope: \( m = -1/2 \). Line: \( y = -0.5x \).
3. Find the intersection: \( 2x + 5 = -0.5x \Rightarrow 2.5x = -5 \Rightarrow x = -2 \).
4. Plug \(x\) into either line: \( y = -0.5(-2) = 1 \).
5. The point is \((-2, 1)\).

Answer: (-2, 1)

Example 14 (MIT)

Three positive integers \(a, b, c\) satisfy \( a < b < c \). Their sum is 20 and their product is 240. Find \(c\).

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1. We have \( a+b+c = 20 \) and \( abc = 240 \).
2. Prime factorization of 240: \( 2^4 \times 3 \times 5 \).
3. Possible triplets that sum to 20:
– Try \( c=12 \): \( a+b=8, ab=20 \). Factors of 20 that sum to 8? No.
– Try \( c=10 \): \( a+b=10, ab=24 \). Factors of 24 are 4 and 6. \( 4+6=10 \).
4. The integers are \( 4, 6, 10 \). They satisfy all conditions.

Answer: 10

Example 15 (MIT)

If \( x + \frac{1}{x} = 5 \), find the value of \( x^2 + \frac{1}{x^2} \).

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1. Take the original equation and square both sides: \( (x + \frac{1}{x})^2 = 5^2 \).
2. Expand the left side: \( x^2 + 2(x)(\frac{1}{x}) + \frac{1}{x^2} = 25 \).
3. Simplify the middle term: \( x^2 + 2 + \frac{1}{x^2} = 25 \).
4. Subtract 2 from both sides: \( x^2 + \frac{1}{x^2} = 23 \).

Answer: 23

Advanced Practice Questions (30) — (Hard to MIT Level)

1. 1. In the system \( \begin{cases} kx – 3y = 4 \\ 4x – 5y = 7 \end{cases} \), for what value of \(k\) will the system have no solution?
   Show Detailed Solution

   For no solution, the lines must be parallel (same slope) but have different y-intercepts.
   1. Slope of eq 1: Isolate \(y \Rightarrow 3y = kx – 4 \Rightarrow y = \frac{k}{3}x – \frac{4}{3}\). So, \(m_1 = \frac{k}{3}\).
   2. Slope of eq 2: Isolate \(y \Rightarrow 5y = 4x – 7 \Rightarrow y = \frac{4}{5}x – \frac{7}{5}\). So, \(m_2 = \frac{4}{5}\).
   3. Set \(m_1 = m_2 \Rightarrow \frac{k}{3} = \frac{4}{5} \Rightarrow k = \frac{12}{5}\) or 2.4.

2. 2. Line \(L\) passes through \((a, 0)\) and \((0, b)\). If \(a+b = 12\) and the slope of the line is \(-3\), find the value of \(a\).
   Show Detailed Solution

   1. Slope formula: \(m = \frac{y_2 – y_1}{x_2 – x_1} \Rightarrow -3 = \frac{b – 0}{0 – a} \Rightarrow -3 = \frac{b}{-a}\).
   2. Simplify: \(3a = b\).
   3. Substitute into the given sum: \(a + (3a) = 12 \Rightarrow 4a = 12 \Rightarrow a = 3\).

3. 3. If \( \frac{x-y}{x+y} = \frac{3}{5} \), what is the value of the expression \( \frac{2x}{y} \)?
   Show Detailed Solution

   1. Cross-multiply: \(5(x – y) = 3(x + y) \Rightarrow 5x – 5y = 3x + 3y\).
   2. Combine like terms: \(2x = 8y \).
   3. Solve for \(x/y\): \( \frac{x}{y} = \frac{8}{2} = 4\).
   4. Target: \( \frac{2x}{y} = 2(4) = 8\).

4. 4. A line with slope \(m\) and y-intercept \(b\) satisfies \(m+b = 10\). If the line passes through the point \((2, 4)\), find the value of \(m\).
   Show Detailed Solution

   1. General equation: \(y = mx + b\).
   2. Plug in the point: \(4 = m(2) + b \Rightarrow 4 = 2m + b\).
   3. From \(m+b=10\), we know \(b = 10 – m\).
   4. Substitute: \(4 = 2m + (10 – m) \Rightarrow 4 = m + 10 \Rightarrow m = -6\).

5. 5. For what value of \(a\) is the line \(ax + 2y = 10\) perpendicular to the line \(3x + (a-1)y = 8\)?
   Show Detailed Solution

   1. Slope of first line: \(m_1 = -\frac{A}{B} = -\frac{a}{2}\).
   2. Slope of second line: \(m_2 = -\frac{3}{a-1}\).
   3. Perpendicular rule (\(m_1 \cdot m_2 = -1\)): \( \left(-\frac{a}{2}\right)\left(-\frac{3}{a-1}\right) = -1 \).
   4. Simplify: \( \frac{3a}{2a – 2} = -1 \Rightarrow 3a = -2a + 2 \Rightarrow 5a = 2 \Rightarrow a = 0.4\).

6. 6. Solve the inequality \( |x – 4| \le |x + 2| \) for \(x\).
   Show Detailed Solution

   1. This asks where the distance to 4 is less than or equal to the distance to -2.
   2. Find the midpoint: \(\frac{4 + (-2)}{2} = 1\).
   3. At \(x=1\), the distances are equal. Test \(x=2\): \(|2-4|=2\), \(|2+2|=4\). Since \(2 \le 4\), the region includes values to the right of 1.
   4. Answer: \(x \ge 1\).

7. 7. Find the sum of all integers \(k\) that satisfy the inequality \( |2k – 7| < 5 \).
   Show Detailed Solution

   1. Open the absolute value: \(-5 < 2k - 7 < 5\).
   2. Add 7: \(2 < 2k < 12\).
   3. Divide by 2: \(1 < k < 6\).
   4. Integers are: \(2, 3, 4, 5\).
   5. Sum: \(2+3+4+5 = 14\).

8. 8. If \( f(x) = \frac{1}{2}x + k \) and \( f(10) = 2 \), find the x-intercept of the graph of \(f\).
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   1. Solve for \(k\): \(2 = \frac{1}{2}(10) + k \Rightarrow 2 = 5 + k \Rightarrow k = -3\).
   2. Find x-intercept (set \(y=0\)): \(0 = \frac{1}{2}x – 3 \Rightarrow \frac{1}{2}x = 3 \Rightarrow x = 6\).

9. 9. For what value of \(c\) does the equation \( |x – 5| + 3 = c \) have exactly one solution?
   Show Detailed Solution

   1. Isolate the absolute value: \(|x – 5| = c – 3\).
   2. An absolute value has exactly one solution only when it equals zero.
   3. Set \(c – 3 = 0 \Rightarrow c = 3\).

10. 10. Solve for \(x\): \( \frac{2x – 3}{5} – \frac{x + 1}{2} > 1 \).
    Show Detailed Solution

    1. Multiply by 10 (LCD): \(2(2x-3) – 5(x+1) > 10\).
    2. Expand: \(4x – 6 – 5x – 5 > 10\).
    3. Simplify: \(-x – 11 > 10 \Rightarrow -x > 21\).
    4. Flip sign: \(x < -21\).

11. 11. The function \( f(x) = ax + b \) satisfies \( f(0) = 2 \) and \( f(f(1)) = 8 \). Find \(a\) if \(a > 0\).
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    1. \(f(0)=2 \Rightarrow b=2\). So \(f(x) = ax+2\).
    2. \(f(1) = a+2\).
    3. \(f(f(1)) = a(a+2) + 2 = 8\).
    4. \(a^2 + 2a – 6 = 0\). Using quadratic formula: \(a = \frac{-2 \pm \sqrt{4 – 4(1)(-6)}}{2} = \frac{-2 \pm \sqrt{28}}{2} = -1 \pm \sqrt{7}\).
    5. Since \(a > 0\), \(a = \sqrt{7} – 1\).

12. 12. Lines \(y=2x\), \(y=-0.5x\), and \(y=k\) form a triangle with area 20. Find \(k\) where \(k > 0\).
    Show Detailed Solution

    1. Vertices: Origin \((0,0)\), intersection of \(y=2x\) and \(y=k\) is \((\frac{k}{2}, k)\), intersection of \(y=-0.5x\) and \(y=k\) is \((-2k, k)\).
    2. Base length: \(\frac{k}{2} – (-2k) = 2.5k\). Height: \(k\).
    3. Area: \(0.5 \cdot 2.5k \cdot k = 1.25k^2 = 20 \Rightarrow k^2 = 16 \Rightarrow k=4\).

13. 13. A chemist has a 10L solution of 20% alcohol. How much 60% solution must be added to reach a 50% concentration?
    Show Detailed Solution

    1. Alcohol balance: \(0.20(10) + 0.60x = 0.50(10 + x)\).
    2. \(2 + 0.6x = 5 + 0.5x \Rightarrow 0.1x = 3 \Rightarrow x = 30\) Liters.

14. 14. Find the perpendicular distance between parallel lines \(y = 3x + 5\) and \(y = 3x – 5\).
    Show Detailed Solution

    1. Distance formula: \(d = \frac{|C_1 – C_2|}{\sqrt{1 + m^2}}\).
    2. \(d = \frac{|5 – (-5)|}{\sqrt{1 + 3^2}} = \frac{10}{\sqrt{10}} = \sqrt{10}\).

15. 15. If \( x^2 – y^2 = 24 \) and \( x + y = 3 \), find the value of \( x – 2y \).
    Show Detailed Solution

    1. \((x+y)(x-y)=24 \Rightarrow 3(x-y)=24 \Rightarrow x-y=8\).
    2. System: \(x+y=3\) and \(x-y=8\). Add them: \(2x=11 \Rightarrow x=5.5, y=-2.5\).
    3. \(x – 2y = 5.5 – 2(-2.5) = 10.5\).

16. 16. A set of \(n\) consecutive even integers has a sum of 0. What is their average?
    Show Detailed Solution

    For any arithmetic sequence, \(Sum = n \times Average\). If the sum is 0 and \(n \neq 0\), the average must be 0.

17. 17. Line \(A\) is \(y=3x+2\). Line \(B\) is its reflection over \(y=x\). What is the slope of Line \(B\)?
    Show Detailed Solution

    Reflection over \(y=x\) is the inverse. The slope of the inverse of a line with slope \(m\) is \(1/m\). Slope = \(1/3\).

18. 18. How many integer pairs \((x, y)\) satisfy \( |x| + |y| \le 2 \)?
    Show Detailed Solution

    List them: \((0,0), (\pm 1, 0), (0, \pm 1), (\pm 2, 0), (0, \pm 2), (\pm 1, \pm 1)\). Total = 13.

19. 19. If \( x+y=7, y+z=10, x+z=13 \), find \( x+y+z \).
    Show Detailed Solution

    Add all: \(2x+2y+2z = 30 \Rightarrow x+y+z=15\).

20. 20. What is the minimum value of the function \( f(x) = |x-1| + |x-5| \)?
    Show Detailed Solution

    This is the sum of distances to 1 and 5. For any \(x\) between 1 and 5, the distance is constant at \(5-1=4\). Outside this range, it increases. Minimum = 4.

21. 21. If \( x + \frac{1}{x} = 3 \), find the value of \( x^4 + \frac{1}{x^4} \).
    Show Detailed Solution

    1. Square: \(x^2 + 2 + 1/x^2 = 9 \Rightarrow x^2 + 1/x^2 = 7\).
    2. Square again: \(x^4 + 2 + 1/x^4 = 49 \Rightarrow x^4 + 1/x^4 = 47\).

22. 22. A car covers half a trip at speed \(v_1\) and the other half at speed \(v_2\). What is the average speed?
    Show Detailed Solution

    \(Avg = \frac{Total Dist}{Total Time} = \frac{D}{\frac{D/2}{v_1} + \frac{D/2}{v_2}} = \frac{2v_1v_2}{v_1 + v_2}\).

23. 23. How many real roots exist for the equation \( |x^2 – 4| = 5 \)?
    Show Detailed Solution

    1. \(x^2 – 4 = 5 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3\).
    2. \(x^2 – 4 = -5 \Rightarrow x^2 = -1\) (No real roots).
    3. Total = 2 real roots.

24. 24. If \( a+b+c=0 \), evaluate \( \frac{a^2}{bc} + \frac{b^2}{ac} + \frac{c^2}{ab} \).
    Show Detailed Solution

    1. Combine: \(\frac{a^3 + b^3 + c^3}{abc}\).
    2. Identity: If \(a+b+c=0\), then \(a^3+b^3+c^3 = 3abc\).
    3. Result: \(3abc/abc = 3\).

25. 25. Runner A (6 m/s) and Runner B (4 m/s) start together on a 400m track in the same direction. When do they meet?
    Show Detailed Solution

    Relative speed = \(6 – 4 = 2\) m/s. They meet when the lead is exactly 1 lap: \(t = 400/2 = 200\) seconds.

26. 26. Solve for \(x\): \( \sqrt{x + \sqrt{x + \sqrt{x…}}} = 5 \).
    Show Detailed Solution

    Substitute the infinite part: \(\sqrt{x + 5} = 5 \Rightarrow x+5=25 \Rightarrow x=20\).

27. 27. If \( f(x) = \frac{x}{x-1} \), find the simplified form of \( f(f(x)) \).
    Show Detailed Solution

    \(f(f(x)) = \frac{x/(x-1)}{x/(x-1) – 1} = \frac{x/(x-1)}{1/(x-1)} = x\).

28. 28. If \( 2^x = 3^y = 6^{-z} \), find \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \).
    Show Detailed Solution

    Set all to \(k\). \(x = \log_2 k\), \(y = \log_3 k\), \(z = \log_{1/6} k\).
    \(1/x + 1/y + 1/z = \log_k 2 + \log_k 3 + \log_k (1/6) = \log_k (2 \cdot 3 \cdot 1/6) = \log_k 1 = 0\).

29. 29. A line passes through \((2, 3)\) such that the area in the first quadrant is minimized. Find its slope.
    Show Detailed Solution

    Minimized area occurs when the given point is the midpoint of the segment between the axes. Intercepts are \((4,0)\) and \((0,6)\). Slope = \(-6/4 = -1.5\).

30. 30. What is the remainder when \( x^{100} – 2x^{51} + 1 \) is divided by \( x – 1 \)?
    Show Detailed Solution

    Remainder Theorem: Plug in \(x=1\). \(1^{100} – 2(1)^{51} + 1 = 1 – 2 + 1 = 0\).

Next Steps: Path to 800

Mastering the “Heart of Algebra” is the foundation of a top-tier SAT score. To push toward an 800, ensure you can translate word problems into equations in under 30 seconds. Are you ready to master Problem Solving & Data Analysis?