SAT Linear Equations in One Variable: Solve It, or Prove It Can't Be Solved

The core idea

Can you divide $4t-24=8$ by 4 before removing the $-24$? Yes—provided you divide every term. It becomes $t-6=2$, so $t=8$. Adding 24 first also works: $4t=32$, then $t=8$. Both routes preserve the same solutions.

A linear equation says that two expressions have equal values. Solving means finding all the values that make that statement true. Adding or subtracting the same quantity on both sides, and multiplying or dividing both sides by the same nonzero constant, preserve the solution set. Multiplying by zero loses information; dividing by zero is not allowed.

You do not move a term by magic. You apply the same operation to both sides, then simplify. Choose the order that makes the next line easier to read.

A short route you can check

For $3(x-3)=18$, dividing both sides by 3 gives $x-3=6$. Add 3 to both sides: $x=9$. Check in the original: $3(9-3)=18$.

When there are several terms, a useful order is: distribute, combine like terms, collect the variable, then divide by its remaining nonzero coefficient. This is a helpful route, not the only legal order. Before dividing, check that the variable has not cancelled.

From a story to an equation

Name the unknown and its units. Write an expression for each quantity, then use the relationship in the story to connect them. A one-time fee is a constant; a rate is multiplied by the amount of time, distance, or number of objects to which it applies.

A landscaper charges a 12-dollar delivery fee plus 4 dollars per shrub. A 40-dollar bill gives $4s+12=40$, so $4s=28$ and $s=7$ shrubs. Check: $4(7)+12=40$.

For a price comparison, write both totals. For an included allowance, count only the excess: if a plan includes $g$ units and usage is $u>g$, the additional charge uses $u-g$, not $u$. State the applicable condition before solving.

Worked examples

1. Match the rate to the units

A repair shop charges a 40-dollar call-out fee plus 30 dollars per half hour. Dev's bill is 190 dollars. How many hours of labor were charged?

Let $n$ count half-hour units. The equation is $40+30n=190$. Subtract 40: $30n=150$. Divide by 30: $n=5$. Convert the requested quantity: five half hours is $5/2=2.5$ hours.

Why this route? Counting half hours matches the stated rate. An equally valid route is to convert the rate first to 60 dollars per hour and solve $40+60h=190$. The answer 5 confuses half hours with hours; 10 converts in the wrong direction. Check the bill: $40+30(5)=190$.

2. Distribute the negative sign

Solve $2(3x+4)-5(x-1)=7$. First predict the second expansion.

Now expand the first group: $2(3x+4)=6x+8$. Combining both expansions gives $6x+8-5x+5=7$, so $x+13=7$ and $x=-6$.

Check: $2(3(-6)+4)=-28$ and $-5(-6-1)=35$; their sum is 7. The tempting answer 4 comes from writing $-5x-5$ instead of $-5x+5$. The answer $-2$ comes from multiplying $3x$ by 2 but leaving the 4 unchanged.

3. Count the second phase only once

A tank starts with 240 gallons. A pump drains 15 gallons per minute for the first four minutes, then 12 gallons per minute until 60 gallons remain. Find the total running time $t$.

The first phase removes $15(4)=60$ gallons. The second phase lasts $t-4$ minutes. Therefore

Distribute: $180-12t+48=60$. Combine: $228-12t=60$. Add $12t$ and subtract 60: $168=12t$, so $t=14$ minutes.

Check: the second phase lasts 10 minutes and removes 120 gallons; $240-60-120=60$. Reporting 10 answers only the second-phase question. Writing $12t$ would apply the second rate to time already counted in the first phase.

Before harder practice

Fractions with nonzero integer denominators can be cleared by multiplying every term on both sides by the least common multiple of those denominators. For $\frac{x-2}{3}+\frac{x+1}{2}=4$, multiply by 6: $2(x-2)+3(x+1)=24$. This becomes $5x-1=24$, so $x=5$; the original checks as $1+3=4$. Keep parentheses around each numerator while distributing.

When an identity contains two unknown constants, compare coefficients and constants separately. For $\frac{ax+b}{2}=x+7$ to hold for every $x$, multiplying by 2 gives $ax+b=2x+14$, so $a=2$ and $b=14$. Neither comparison can replace the other.

For successive removals, name the remainder before using the next fraction. If a first removal takes one-quarter of a total $L$, plus 6 units, the remainder is $R=L-(L/4+6)$. If half that remainder is then removed, the final amount is $R/2$, not $L/2$. Substitute the expression for $R$ only after the story's relationships are clear.