ONE-, TWO- & MULTI-STEP EQUATIONS
Solve. x − 7 = 12
x= 19
Solve for r. d = rt
r= d/t
Write the equation made by the cross products (do not solve). 3/5 = x/20
5x = 60
Evaluate. |−8|
8
Solve. x + 5 < 12
x<7
Solve. 3x + 4 = 22
x=6
Solve for b. y = mx + b
b = y − mx
Solve. 4/7 = x/21
x=12
Solve. |x − 4| = 9
x = 13 or x = −5
Solve. 2x − 3 ≥ 9
x ≥ 6
Solve. 4x + 7 = 2x + 19
x=6
Solve for w.
P = 2l + 2w
w = (P − 2l) / 2
Solve. 5/8 = 15/x
x=24
Solve. 2|x + 1| = 10
x = 4 or x = −6 (isolate first: |x + 1| = 5)
Solve. −4x ≥ 20
x ≤ −5 (divide by −4, so the sign flips)
Solve, then name the solution type.
4x + 3 = 4x + 9
No solution.
The variable cancels and 3 = 9 is false.
Solve for h. A = ½ bh
h = 2A / b
Solve. (x − 2)/4 = 9/6
x = 8
6(x − 2) = 36 →
x − 2 = 6
Solve and explain. |x + 5| = −2
No solution. Absolute value is a distance from zero, so it can never equal a negative number.
Solve the compound inequality. −1 < x + 2 ≤ 6
−3 < x ≤ 4
Error analysis.
A student solved 2x − 5 = 15 and wrote 2x = 10, so x = 5.
Find the mistake and solve correctly.
Correct: 2x = 20, so x = 10.
Solving P = 2l + 2w for w, a student wrote w = P/2 − 2l. What is the mistake and the correct answer?
They divided only part of the equation by 2. Subtract 2l first, then divide everything by 2: w = (P − 2l) / 2 .
Error analysis. A student solved (x + 1)/3 = 8/12 and wrote
12(x + 1) = 24 → 12x + 1 = 24.
Find the mistake and solve correctly.
They didn't distribute the 12 to BOTH terms.
12(x + 1) =
12x + 12 = 24 →
12x = 12 →
x = 1
Error analysis. A student solved |x − 5| = 8 and wrote x − 5 = 8, so x = 13. Find the mistake and give the full solution.
They only solved one case. Absolute value splits into two: x − 5 = 8 (x = 13) and x − 5 = −8 (x = −3). Solutions: x = 13 or x = −3.
Error analysis. A student solved −3x + 2 ≤ 14 and wrote −3x ≤ 12 → x ≤ −4.
Find the mistake and solve correctly.
Dividing by a negative flips the sign. −3x ≤ 12 → x ≥ −4.