Math – Algebra

Explore expert-written guides that decode the mindset, strategy, and science behind SAT success.

Heart of Algebra — SAT Topic Page

Focus on linear relationships: equations, systems, inequalities, and modeling. Use the worked examples to learn methods and the practice set for timed accuracy.

Key Forms & Concepts (Quick Reference)

FormTemplateNotes
Slope \( m = \dfrac{y_2 – y_1}{x_2 – x_1} \) Rate of change per unit x
Slope–Intercept \( y = mx + b \) Intercept \(b\) at \((0,b)\)
Point–Slope \( y – y_1 = m(x – x_1) \) Great for fast equation building
Standard \( Ax + By = C \) Good for elimination
System \( \begin{cases} a_1x+b_1y=c_1\\ a_2x+b_2y=c_2\end{cases} \) Substitution / elimination
Inequality \( Ax + By \le C \) Graph uses shading

Worked Examples (10) — (Difficulty level)

Linear Equations

Example 1 (Medium)

Solve \( \dfrac{3x-5}{4} – \dfrac{x+1}{2} = \dfrac{7}{4} \).

Show solution

Multiply both sides by 4: \( (3x-5) – 2(x+1) = 7 \Rightarrow 3x-5 -2x -2 = 7 \Rightarrow x -7 = 7 \Rightarrow x = 14 \).

Answer: \( x=14 \)

Example 2 (Medium)

Find the equation of the line through \((2,-3)\) and \((-4,9)\).

Show solution

\( m=\dfrac{9-(-3)}{-4-2} = -2 \). Using point–slope with \((2,-3)\): \( y+3 = -2(x-2) \Rightarrow y = -2x + 1 \).

Answer: \( y = -2x + 1 \)

Systems of Equations

Example 3 (Ivy)

For which real \(k\) does the system \( \begin{cases} 2x + ky = 6 \\ 4x + 10y = 18 \end{cases} \) have infinitely many solutions?

Show solution

Reduce second: \(2x+5y=9\). For identical equations, need \(k=5\) and RHS equal: but \(6\ne9\). Thus no \(k\) gives identity ⇒ no value produces infinitely many solutions.

Answer: No real \(k\).

Example 4 (Ivy)

Find \(k\) such that the system \( \begin{cases} x + 2y = 7 \\ kx + 2ky = 14 \end{cases} \) has no solution.

Show solution

If \(k\ne0\), second is \(x+2y=\tfrac{14}{k}\). For no solution, require \(\tfrac{14}{k}\ne7\Rightarrow k\ne2\) and \(k\ne0\).

Answer: Any \(k\in\mathbb{R}\setminus\{0,2\}\).

Inequalities

Example 5 (Medium)

Solve \( -3(2x-5) + 7 \le 4x – 1 \).

Show solution

Expand: \( -6x+22 \le 4x-1 \Rightarrow 23 \le 10x \Rightarrow x \ge \dfrac{23}{10} \).

Answer: \( x \ge 2.3 \)

Example 6 (Ivy)

Solve \( |2x-7| < 5 - x \).

Show solution

Need \(x<5\). Case split: If \(x\ge3.5\) ⇒ \(2x-7 < 5-x\Rightarrow x<4\). If \(x<3.5\) ⇒ \(-(2x-7) < 5-x\Rightarrow x>2\). Combine ⇒ \((2,4)\).

Answer: \( (2,4) \)

Word & Modeling

Example 7 (Ivy)

A plan charges base \(b\) and \(c\) dollars per GB. Bills: \$29 for 3 GB and \$53 for 7 GB. Find \(b,c\).

Show solution

System: \( b+3c=29,\; b+7c=53 \Rightarrow c=6,\; b=11 \).

Answer: \( b=11,\; c=6 \).

Example 8 (MIT)

Two trains: A at 60 mph from \(t=0\); B at \(v\) mph starting at \(t=0.5\) hr. At \(t=2.5\) hr they meet. Find \(v\).

Show solution

A distance: \(60t\). B runs for \(t-0.5\): \(v(t-0.5)\). At \(t=2.5\): \(150 = 2v + 30\Rightarrow v=60\).

Answer: \( v=60 \) mph.

Example 9 (MIT)

Almonds \$7/kg and peanuts \$3/kg to make 10 kg at \$5.20/kg. Find amounts.

Show solution

\(a+p=10,\;7a+3p=52\Rightarrow a=5.5,\;p=4.5\).

Answer: Almonds 5.5 kg; peanuts 4.5 kg.

Example 10 (MIT)

Line through \((t,2t-1)\) and \((3,1)\) has slope \(-\dfrac{1}{2}\). Find \(t\).

Show solution

\( -\tfrac{1}{2} = \dfrac{2-2t}{3-t} \Rightarrow t=\tfrac{5}{3} \).

Answer: \( t=\tfrac{5}{3} \).

Practice Questions (20)

Linear Equations
  1. \( 5(2x-3) – 3(x+4) = 7 \).
    Solution

    \(10x-15-3x-12=7\Rightarrow 7x-27=7\Rightarrow x= \dfrac{34}{7} \).

  2. Through \((1,4)\) slope \(-3\).
    Solution

    \(y-4=-3(x-1)\Rightarrow y=-3x+7\).

  3. Through \((2,5)\), \((6,-3)\).
    Solution

    \(m=-2\Rightarrow y=-2x+9\).

  4. Parallel to \(2y=-6x+1\). Find \(k\) in \(y=kx+4\).
    Solution

    slope \(-3\) ⇒ \(k=-3\).

  5. Find \(b\): line slope 2 passes through \((3,11)\).
    Solution

    \(11=6+b\Rightarrow b=5\).

Systems of Equations
  1. \( \{\,3x+y=11,\;2x-3y=1\,\} \).
    Solution

    \(x=\tfrac{34}{11},\;y=-\tfrac{47}{11}\).

  2. \( \{\,x+4y=13,\;5x-2y=9\,\} \).
    Solution

    \(y=\tfrac{28}{11},\;x=\tfrac{75}{11}\).

  3. No solution for \( \{\,x+2y=5,\;kx+2ky=9\,\} \) when?
    Solution

    \(k\ne0,\;k\ne9/5\).

  4. \( \{\,2x-5y=7,\;4x-10y=14\,\} \).
    Solution

    Infinitely many (same line).

  5. \( \{\,7x+3y=1,\;5x-4y=29\,\} \).
    Solution

    \(x=\tfrac{91}{43},\;y=-\tfrac{198}{43}\).

Inequalities
  1. \( 4x – 7 \le 3x + 2 \).
    Solution

    \(x\le9\).

  2. \( -2(3-x) > 7 – x \).
    Solution

    \(x>\tfrac{13}{3}\).

  3. \( |x-5| \ge 4 \).
    Solution

    \(x\le1\) or \(x\ge9\).

  4. \( |3x+1| < 2x+7 \).
    Solution

    \((-\tfrac{8}{5},6)\).

  5. Solution set of \( x+b>2b-3 \) is all \(x>k\). Find \(k\).
    Solution

    \(k=b-3\).

Word & Modeling
  1. Taxi: \(C=4+2.4m\). Miles for \$28?
    Solution

    \(m=10\).

  2. Tank: net 21 L/hr if leak 3 L/hr and fill \(r\) L/hr.
    Solution

    \(r=24\).

  3. Break-even: \(R=px\), \(C=120+4x\), \(x=40\).
    Solution

    \(p=7\).

  4. Mixture: 40% with 10% to get 25%, total 30 L.
    Solution

    \(a=b=15\).

  5. Decay: \(y=5-0.8t\). When \(y=0\)?
    Solution

    \(t=6.25\) days.

Next Steps

Repeat methods under time. Target: medium \(<40\)s; advanced \(<90\)s; accuracy \(\ge 80\%\). Then continue to Problem Solving & Data Analysis.