Algebra I Practice Exam I Study Guide

Algebra I Practice Exam I

I. Multi-Step Equations (4 Questions)

  • Question 1: Solve for $x$ in the equation:

    • 3(x−5)+2=173(x - 5) + 2 = 17
    • Solution Steps:
    1. Distribute: 3x−15+2=173x - 15 + 2 = 17
    2. Combine like terms: 3x−13=173x - 13 = 17
    3. Add 13 to both sides: 3x=303x = 30
    4. Divide by 3: x=10x = 10

  • Question 2: Solve for $x$ in the equation:

    • 8x−(2x+4)=208x - (2x + 4) = 20
    • Solution Steps:
    1. Distribute: 8x−2x−4=208x - 2x - 4 = 20
    2. Combine like terms: 6x−4=206x - 4 = 20
    3. Add 4 to both sides: 6x=246x = 24
    4. Divide by 6: x=4x = 4

  • Question 3: Solve for $x$ in the equation:

    • 23x+5=13\frac{2}{3}x + 5 = 13
    • Solution Steps:
    1. Subtract 5 from both sides: 23x=8\frac{2}{3}x = 8
    2. Multiply both sides by 32\frac{3}{2}: x=12x = 12

  • Question 4: Solve for $x$ in the equation:

    • 5(2x−1)=3x+305(2x - 1) = 3x + 30
    • Solution Steps:
    1. Distribute: 10x−5=3x+3010x - 5 = 3x + 30
    2. Subtract 3x from both sides: 7x−5=307x - 5 = 30
    3. Add 5 to both sides: 7x=357x = 35
    4. Divide by 7: x=5x = 5

II. Proportions (2 Questions)

  • Question 5: Solve for $x$ in the proportion:

    • x6=512\frac{x}{6} = \frac{5}{12}
    • Solution Steps:
    1. Cross-multiply: 12x=3012x = 30
    2. Divide by 12: x=3012=2.5x = \frac{30}{12} = 2.5

  • Question 6: Solve for $x$ in the proportion:

    • x+34=92\frac{x + 3}{4} = \frac{9}{2}
    • Solution Steps:
    1. Cross-multiply: 2(x+3)=362(x + 3) = 36
    2. Divide by 2: x+3=18x + 3 = 18
    3. Subtract 3 from both sides: x=15x = 15

III. Absolute Value Equations (4 Questions)

  • Question 7: Solve for $x$ in the equation:

    • ∣x−4∣=10|x - 4| = 10
    • Solution Steps:
    1. Set up two equations:
      • Case 1: x - 4 = 10
        ightarrow x = 14
      • Case 2: x - 4 = -10
        ightarrow x = -6

  • Question 8: Solve for $x$ in the equation:

    • 2∣3x+1∣=142|3x + 1| = 14
    • Solution Steps:
    1. Divide both sides by 2: ∣3x+1∣=7|3x + 1| = 7
    2. Set up two equations:
      • Case 1: 3x + 1 = 7
        ightarrow 3x = 6
        ightarrow x = 2
      • Case 2: 3x + 1 = -7
        ightarrow 3x = -8
        ightarrow x = -\frac{8}{3}

  • Question 9: Solve for $x$ in the equation:

    • ∣x2−5∣+3=8\left|\frac{x}{2} - 5\right| + 3 = 8
    • Solution Steps:
    1. Subtract 3 from both sides: ∣x2−5∣=5\left|\frac{x}{2} - 5\right| = 5
    2. Set up two equations:
      • Case 1: \frac{x}{2} - 5 = 5
        ightarrow \frac{x}{2} = 10
        ightarrow x = 20
      • Case 2: \frac{x}{2} - 5 = -5
        ightarrow \frac{x}{2} = 0
        ightarrow x = 0

  • Question 10: Solve for $x$ in the equation:

    • ∣2x+7∣=15|2x + 7| = 15
    • Solution Steps:
    1. Set up two equations:
      • Case 1: 2x + 7 = 15
        ightarrow 2x = 8
        ightarrow x = 4
      • Case 2: 2x + 7 = -15
        ightarrow 2x = -22
        ightarrow x = -11

IV. Literal Equations (2 Questions)

  • Question 11: Solve for $h$ in the equation:

    • V=lwhV = lwh
    • Solution Steps:
    1. Isolate $h$: h=Vlwh = \frac{V}{lw}

  • Question 12: Solve for $y$ in the equation:

    • 2x+3y=122x + 3y = 12
    • Solution Steps:
    1. Isolate $y$: 3y=12−2x3y = 12 - 2x
    2. Divide by 3: y=12−2x3y = \frac{12 - 2x}{3}

V. Word Problems (4 Questions)

  • Question 13: Geometry Problem:

    • Problem Statement: The length of a rectangle is 5 cm more than its width. If the perimeter is 50 cm, find the dimensions (length and width).
    • Solution Steps:
    1. Let width = $w$ cm.
    2. Length = $w + 5$ cm.
    3. Perimeter formula: P=2(l+w)P = 2(l + w)
    4. Setup equation: 50=2((w+5)+w)50 = 2((w + 5) + w)
    5. Simplifying: 50 = 2(2w + 5)
      ightarrow 50 = 4w + 10
    6. Solving yields: 4w = 40
      ightarrow w = 10 (Width)
    7. Length = $15$ cm.

  • Question 14: Numbers Problem (Practice A):

    • Problem Statement: The larger of two numbers is 5 more than three times the smaller number. If the sum of the two numbers is 65, find both numbers.
    • Solution Steps:
    1. Let smaller number = $x$. Larger number = $3x + 5$.
    2. Setup equation: x+(3x+5)=65x + (3x + 5) = 65
    3. Simplifying: 4x + 5 = 65
      ightarrow 4x = 60
      ightarrow x = 15 (Smaller Number)
    4. Larger number = $50$.

  • Question 15: Numbers Problem (Practice B):

    • Problem Statement: One number is 8 less than twice another number. If their sum is 22, find both numbers.
    • Solution Steps:
    1. Let the first number = $x$ and the second = $y$,
    2. Setup equations: y=2x−8y = 2x - 8 and x+y=22x + y = 22
    3. Substitute: x+(2x−8)=22x + (2x - 8) = 22
    4. Solving yields: 3x - 8 = 22
      ightarrow 3x = 30
      ightarrow x = 10 (First Number)
    5. Second number = $12$.

  • Question 16: Cost Problem:

    • Problem Statement: A taxi service charges a flat fee of $5 plus $2 for every mile traveled. If the total bill was $23, how many miles did the taxi travel?
    • Solution Steps:
    1. Let number of miles = $m$.
    2. Setup equation: 5+2m=235 + 2m = 23
    3. Solve for $m$: 2m = 18
      ightarrow m = 9 (Miles traveled).