ACT Math study guide · Chapter 45
Backsolving, Plugging In, and Estimating
The previous chapter covered which tool to use once a route is chosen. This chapter covers a different decision: whether to solve a question directly at all. Every ACT Math question has four answer choices, and that structure makes three alternate methods genuinely useful, not just a fallback for when algebra fails.
Backsolving means testing an answer choice inside the condition the question describes, instead of solving for the unknown symbolically. It works well when the condition is easy to check but awkward to solve directly, and when the answer choices are specific numbers instead of expressions.
Worked Example A rectangle's length is 4 more than its width. If the area is 96, find the width.
Solving this directly means factoring w²+4w−96=0, which is not obviously factorable at a glance. Testing each width instead means one multiplication per choice.
A) 6: 6×10=60.
B) 8: 8×12=96.
C) 10: 10×14=140.
D) 12: 12×16=192.
Only choice B satisfies the condition, so the width is 8.
Plugging in convenient values means choosing a specific number for a variable that appears in every answer choice, instead of manipulating the general expression. It works well on “which expression is equivalent” questions, where the answer choices are themselves expressions in the same variable.
Only the choice x matches when x=4, so x is the surviving answer choice in this multiple-choice check.
Estimating means using a bound or an approximation to eliminate answer choices, instead of computing an exact value. It works well when the answer choices are spread far apart, or when a quick approximation of a formula is enough to identify the one choice in range.
Among widely spaced choices, 113 identifies the correct one without a precise calculation.
None of these three methods is always faster. A one-step equation is faster to solve directly than to backsolve, since direct solving is a single calculation and backsolving is up to four. An expression with no variable to substitute cannot use plugging in. A question that asks for an exact value, not an approximate one, cannot be finished by estimating alone. Before choosing a method, check whether the answer choices are numbers or expressions, how many steps direct solving would take, and whether an approximate answer is actually sufficient for what the question asks.
Plugging in is an elimination method, not a proof of identity. A choice that fails for one legal input is eliminated immediately. A choice that works for one input has only passed that test; another non-equivalent expression can happen to match there. Use a second strategically chosen input when needed. If more than one choice still survives, finish with algebra or another decisive check instead of treating two matching values as proof that the expressions are identical.
▸ Practice
Question 1
- AA. 6
- BB. 8
- CC. 10
- DD. 12
Question 2
- FF. 0
- GG. 2
- HH. 4
- JJ. 5
Question 3
- AA. x^(1/2)
- BB. x
- CC. x^(3/2)
- DD. x²
Question 4
- FF. 126
- GG. 251
- HH. 502
- JJ. 1005
Question 5
- AA. 1
- BB. 3
- CC. 6
- DD. 9
Question 6
- FF. 2
- GG. 2^n
- HH. 4^n
- JJ. 2^(n+1)
Question 7
- AA. 7
- BB. 10
- CC. 14
- DD. 20
Question 8
- FF. 35
- GG. 45
- HH. 48
- JJ. 50
Question 9
- AA. x²+4x+3
- BB. x²+3
- CC. x²+4x+1
- DD. x²+1
Question 10
- FF. 8
- GG. 11
- HH. 13
- JJ. 16
Question 11
- AA. −9
- BB. 2
- CC. 5
- DD. 7
Question 12
- FF. x
- GG. x²
- HH. x³
- JJ. 2x
▸ Answers and Explanations
Show the answers and explanations
1: B. Backsolve: test each width in w(w+4). A) 6(10)=60. B) 8(12)=96. C) 10(14)=140. D) 12(16)=192. Only B matches the given area of 96; solving w²+4w−96=0 directly gives the same width but requires the quadratic formula first.
2: G. Backsolve: substitute each choice into x³−6x²+11x−6. F) 0: −6. G) 2: 8−24+22−6=0. H) 4: 64−96+44−6=6. J) 5: 125−150+55−6=24. Only G satisfies the equation.
3: B. Plug in x=4: (4^0.5·4^1.5)/4=(2×8)/4=4. Check each choice at x=4: A) 4^0.5=2. B) 4. C) 4^1.5=8. D) 4²=16. Only B matches; the expression simplifies to x.
4: H. Estimate using π≈3.14: volume=π(4)²(10)=160π≈160(3.14)=502.4. Only H is close to this value; the other choices are not within a reasonable rounding distance.
5: C. Backsolve to avoid checking for extraneous roots separately: A) x=1: √4=2, and 1−3=−2; 2≠−2. B) x=3: √6≈2.449, and 3−3=0; not equal. C) x=6: √9=3, and 6−3=3; equal. D) x=9: √12≈3.464, and 9−3=6; not equal. Only C satisfies the equation as written, which also confirms that solving algebraically and squaring both sides would have introduced x=1 as an extraneous root.
6: F. Plug in n=3: (2³·2⁴)/2⁶=(8×16)/64=2. Check each choice: F) 2. G) 2³=8. H) 4³=64. J) 2⁴=16. Only F matches; the expression simplifies to the constant 2 regardless of n, since 2^(2n+1)/2^(2n)=2.
7: C. Estimate each root: √50≈7.07 and √48≈6.93, so the sum is close to 14. Only C is near this value; the other choices are far enough away that an exact calculation is not needed to eliminate them.
8: H. Backsolve using the multiplier 1.25: F) 35(1.25)=43.75. G) 45(1.25)=56.25. H) 48(1.25)=60. J) 50(1.25)=62.5. Only H produces the given result of 60.
9: A. Plug in x=1: g(1)=1+2=3, and f(3)=3²−1=8. Check each choice at x=1: A) 1+4+3=8. B) 1+3=4. C) 1+4+1=6. D) −1+2=1. Only A matches; expanding (x+2)²−1 directly gives the same x²+4x+3.
10: H. Estimate instead of computing √176 exactly: 13²=169 and 14²=196, and 176 sits closer to 169, so the root is a little above 13. Only H is consistent with that estimate; the exact value is about 13.27.
11: D. Backsolve instead of tracking both cases of the absolute value: A) |2(−9)−5|=23. B) |2(2)−5|=1. C) |2(5)−5|=5. D) |2(7)−5|=9. Only D satisfies the equation.
12: G. Testing two different negative integers illustrates the pattern before relying on the general reason. At x=−2: F) −2. G) 4. H) −8. J) −4. At x=−5: F) −5. G) 25. H) −125. J) −10. Only x² is positive in both cases, and this holds for every negative x, since squaring any nonzero real number removes its sign.