1
Kill the denominators in one move.
Multiply every term on both sides by the LCD — including the terms with no fraction at all.
Denominators
3
2
LCD
6
6 ·
2x − 53
+ 6 · 4 =
6 ·
x + 72
6 ÷ 3 = 2, so the first term becomes 2(2x − 5)
divide before you multiply
6 · 4 = 24
the plain number counts too
6 ÷ 2 = 3, so the right side becomes 3(x + 7)
one bracket, not one term
2(2x − 5) + 24 = 3(x + 7)
The term without a fraction is the trap
Every SAT item of this shape hides one plain number. Skip it and the
equation stops being true — that is exactly how the answer 27 gets on
the list.
2
Distribute before you collect anything.
The number outside a bracket reaches every term inside it.
2(2x − 5) + 24 = 3(x + 7)
from step 1
4x − 10 + 24 = 3x + 21
2 · 2x, 2 · (−5), 3 · x, 3 · 7
4x + 14 = 3x + 21
−10 + 24 = 14
Combine the loose numbers on each side before you start
moving things across the equals sign. One tidy line beats three untidy ones.
3
Send the variable to one side, the numbers to the other.
Move the smaller x term so the one you keep stays positive.
− 3x both sides− 14 both sides
4x + 14 = 3x + 21
x lives on both sides
x = 7
Desmos is on your screen — use it
The Digital SAT gives you a graphing calculator the whole time. Type the
left side as y = 2x − 53 + 4 and the right side as y = x + 72, then
read the x of the intersection point: 7. Solve by hand when it is quick,
and let the calculator confirm.
4
Read what the question actually asks for.
The SAT often wants an expression, not x — and the expression is usually faster.
3x − 8 = 19, find 3x + 6
3x = 27
3x + 6 = 33
x itself is never needed.
4x + 7 = 31, find 12x + 7
4x = 24
12x = 3 · 24 = 72
12x + 7 = 79
Solving for x would give x = 6 and the same 79 — three extra lines for the same answer.
The favorite distractor is x
When the stem ends "what is the value of 12x + 7", the value of x (here 6)
is sitting in the choices waiting for you. Underline the quantity asked for
before you start.
5
Turn the sentence into the equation, one phrase at a time.
A fixed charge is a plain number; a "per" charge is a coefficient.
| The words | The algebra |
|---|
| a one-time joining fee of 45 dollars | 45 |
| 22 dollars for each month, m | 22m |
| the member paid 265 dollars in all | = 265 |
45 + 22m = 265
the equation the SAT wants
Two plans compared? Set the two expressions equal to each other.
80 + 12d = 30 + 22d
80 dollars + 12 a day vs 30 dollars + 22 a day
50 = 10d
− 30 and − 12d both sides
At 5 days both plans cost 140 dollars: 80 + 60 and 30 + 110. "The same
cost" always means one equation, not two.
6
"Which equation gives r in terms of the other variables?"
Same four moves. Treat every other letter as if it were a number.
A = P(1 + rt)
r is inside the bracket
A = P + Prt
distribute the P
A − P = Prt
− P both sides
r = A − PPt
÷ Pt both sides
r = A − PPt
If the target letter appears twice, factor it out
From ax + 5 = bx − 3, gather first: ax − bx = −8, then factor: x(a − b) = −8,
so x = −8a − b. You can never divide by x while x is still on both sides.
✓
Check: put the answer back in.
Ten seconds of substitution catches every sign slip.
x = 7, left side
14 − 53 + 4 = 3 + 4 = 7
the opening problem
x = 7, right side
7 + 72 = 7
both sides agree ✓
m = 10
45 + 22(10) = 265
the membership total ✓
formula test
P = 500, r = 0.04, t = 3 → A = 500(1.12) = 560
run the original
our rearrangement
560 − 500500 · 3 = 601500 = 0.04
back to r ✓
Remember the flow
Clear · Distribute · Collect · Divide · Re-read the question
A linear equation has exactly one solution unless both sides collapse to the same thing.
