Solve for the x: How to Get x Alone in Any Equation

An equation says two things are equal. The target: a solution is a value you put in place of $x$ that makes the equation true. Take $x-5=4$. Put in 9: $9-5=4$, true. So $x=9$ is the solution. Put in 8: $3=4$, false.

To solve for the x means to find that value without guessing. You rearrange the equation until $x$ stands alone on one side. Ray's 1866 New Higher Algebra adds a comfort: a simple equation has but one root. When you solve for the x on a linear equation, exactly one number is waiting.

The prompt may read how to solve for x, how to solve an equation, or plain solve the equation. The job is one: a value that makes the two sides agree. Every method below is one more way to solve for the x, dressed for a different equation.

Two keys open most boxes: undo an addition, or undo a multiplication. Those two keys solve for the x in most boxes.

Treat an equation like a puzzle. Legal moves: add or subtract the same value on both sides, divide every term by one nonzero value, clear fractions, combine like terms, expand. Here they are, arranged into the routine you run each time you solve for the x:

1. Simplify - expand brackets and clear fractions first, and you solve for the x on clean ground;
2. Collect - move every $x$-term to one side and every plain number to the other, flipping signs. Todhunter's 1889 Algebra calls this move transposition, and Ray's 1866 rule agrees: a term crosses the equals sign only with its sign changed;
3. Reduce - combine like terms on each side;
4. Divide - divide by the coefficient of $x$, leaving it alone.

A solve for the x on a fresh shape starts by reading which operations bury the letter. Then check, always: substitute your answer back into the original equation. There is never any reason to be unsure, because you can always check. The four steps solve for the x; the check certifies it. Run them twice and they run themselves - from then on, every time you solve for the x, muscle memory does the steering. The same backbone carries the how to solve for x drill and every how to solve an equation lesson you have seen: the shapes differ, the routine does not.

Problem. Solve for the x: $2x+7=19$ - addition and multiplication both bury the letter here.

Step 1, undo the +7. Subtract 7 from both sides: $2x=12$ - undo additions first whenever you solve for the x.

Step 2, undo the 2. Divide by 2 to solve for the x: $x=6$.

Check: $2\times6+7=19$, that is $19=19$. True, so $x=6$ stands.

Adapted from Wentworth's 1898 exercise $5x-4=16$ - same shape, new numbers, and the same way to solve for the x. To solve for the x when two operations bury the letter, peel the last one first: that is the whole craft.

Problem. Solve for the x: $5x=2x+12$ - the letter sits on both sides of the equals sign.

Step 1, collect. Subtract $2x$ from both sides: $3x=12$. Collecting comes first whenever you solve for the x on both sides.

Step 2, divide. Divide by 3 to solve for the x: $x=4$.

Check: $5\times4=20$ and $2\times4+12=20$. Both sides give 20, so $x=4$ checks out.

The shape follows a drill ($5x=2x+9$) with new numbers. One collection, one division - to solve for the x, that is the whole job here. The same pattern runs whenever you solve for the x with letters on both sides.

Problem. Solve for the x: $7x-(2x+8)=27$ - a bracket with a minus in front, the classic trap.

Step 1, distribute. The minus hits the whole bracket: $7x-2x-8=27$, so $5x-8=27$. Bracket dissolved - from here, it is Example 1 again.

Step 2, transpose. Add 8 to both sides: $5x=35$.

Step 3, divide. Both sides by 5: $x=7$.

Check: $7\times7-(2\times7+8)=49-22=27$. Both sides agree, so the solve for the x checks out.

Parentheses change the opening, never the method: clear the bracket, then solve for the x exactly as in Example 2. Expand early, and to solve for the x stays a short job.

Problem. Solve for the x: $9-\frac{4}{x}=3+\frac{8}{x}$, where $x\neq0$.

The $x$ sits downstairs. The fix: multiply both sides by $x$ - the entire side, every term, not just the fractions.

Step 1, clear. $9x-4=3x+8$. Fractions gone - one multiplication, and you solve for the x across a fraction bar.

Step 2, collect. $9x-3x=8+4$, so $6x=12$.

Step 3, divide. Divide by 6 to solve for the x: $x=2$.

Check: $9-\frac{4}{2}=7$ and $3+\frac{8}{2}=7$. Both sides give 7, so $x=2$ checks out - and $x=2$ is allowed, since it is not 0.

To solve for the x in fractions: clear first, solve second, every time. From there you solve for the x with the ordinary four steps.

The letter never matters. To solve for w, solve for u, or solve for m is one skill with a new name tag. The moves are the exact moves that solve for the x. Learn them once on $x$ and you own every letter at once: solve for the x or solve for w alike. Ray's 1866 New Higher Algebra lists the four ways a letter gets buried, and the four escapes. Addition, as in $a+u=b$? Subtract: $u=b-a$. Subtraction, $u-a=b$? Add: $u=b+a$. Multiplication, $au=b$? Divide: $u=\frac{b}{a}$. Division, $\frac{u}{a}=b$? Multiply: $u=ab$.

Two mindset rules before the examples. First, in a formula like $2w+5v=20$, treat the target letter as the only unknown and every other letter as a known number - build this habit early. When you solve for the x inside a formula later, the same rule applies. Second, when several letters are candidates, start from the one whose coefficient is already 1 - the easy door first. This section puts solve for the x in three other costumes - w, m, and b. Whether the worksheet says solve for w, solve for u, or solve for b, you run the same routine that lets you solve for the x. The letter is a label, nothing more; one transposition-plus-division finishes every version of the task.

Problem. Solve for w: $7w-6=30-2w$ - the letter on both sides, the same setup you meet when you solve for the x.

Step 1, transpose. Move $-6$ right and $-2w$ left, flipping both signs: $7w+2w=30+6$.

Step 2, reduce. $9w=36$.

Step 3, divide. Both sides by 9: $w=4$ - three moves, the same three that solve for the x.

Check: $7\times4-6=22$ and $30-2\times4=22$. True, so $w=4$ stands.

A warm-up with the variable on both sides - new letter, new numbers. Transpose first: that is how you solve for the x when the letter appears twice. The steps never read the name tag.

Problem. Solve for m: $4(m+3)-2=2(m+7)$.

Step 1, distribute. $4m+12-2=2m+14$. Brackets gone - from here you solve for the m the same way you solve for the x.

Step 2, transpose and reduce. $4m-2m=14-12+2$, so $2m=4$.

Step 3, divide. Both sides by 2: $m=2$.

Check: $4(2+3)-2=18$ and $2(2+7)=18$ - balanced, so $m=2$ stands.

Ray's 1866 exercise $5(x+1)-2=3(x+5)$ has the same shape. Clear the brackets, and you solve for the x on autopilot; any letter inside brackets takes the same road. To solve for the x is to solve for m - nothing else changed.

Problem. A solve for the x under a fraction bar: solve for b in $Q=\frac{c+b}{2}$, then find $b$ when $Q=12$ and $c=6$.

The letter hides under a denominator - Ray's fourth case. Escape by multiplication.

Step 1. Multiply both sides by 2: $2Q=c+b$.

Step 2. Transpose: $b=2Q-c$ - a formula, the usual reward for solving in letters.

Numbers in. $b=2\times12-6=18$. You solve for the x twice here: once in letters, once in numbers.

Check: $\frac{6+18}{2}=12=Q$. The bar is gone; from there you solve for the x the plain way.

This shape is taught with $Q=\frac{c+d}{2}$ solved for $d$ - same two moves, new letters. To solve for the x is to solve for anything; the next section turns from the letter to the wording.

Worksheets and search bars phrase one job a dozen ways: solve x for a number, how can you solve for x, equation finding x. All of it is how to solve for the x in costume. "Solve the equation." and "solve each equation" are repeat instructions, not new methods: run the routine on every item, check included. "Solve x for 12" means make the statement true and report the value. "How to find x" usually shows up in word problems. The only new step is translating the sentence into an equation first and naming the unknown $x$. Then you solve for the x exactly as always.

One prompt adds a demand: "solve algebraically for all values of x" - a solve for the x that owes two answers. When $x$ is squared, two numbers can both work, so the answer lists every root; Example 8 shows it. Another quirk is the colon, as in "solve for y : $6x+2y=16$" - pure punctuation between instruction and equation; isolate $y$ and move on. Which phrasing you started from never changes how you solve for the x: rearrange, isolate, check. To solve for the x under any label is the same four steps. So when a test asks you to solve x for a value, answer it with a full solve for the x on the page.

Problem. Solve algebraically for all values of $x$: $x^2-5x=14$.

When you solve for the x and the letter carries a square, collect everything on one side: $x^2-5x-14=0$. A zero unlocks factoring.

Step 1, factor. Two numbers multiplying to $-14$ and adding to $-5$: $-7$ and 2. So $(x-7)(x+2)=0$.

Step 2, zero product. If two factors multiply to zero, one of them is: $x-7=0$ or $x+2=0$.

Step 3, solve each. $x=7$ or $x=-2$ - both values, no exceptions. A squared letter turns one solve for the x into two.

Check: $7^2-5\times7=14$ and $(-2)^2-5\times(-2)=14$. Both roots hold; a solve for the x with a square owes proof for each.

If a root must also survive a square root, see the companion page on solving where x is a real number. Here, plain algebra settles everything - to solve for the x all the way means factoring first, and every root gets reported.

1. Multiplying only part of a side. To clear $9-\frac{4}{x}=3+\frac{8}{x}$, the slip multiplies just the fractions by $x$. Reminder: hit the entire side, or the sides stop being equal. Every term rides along when you solve for the x.

2. Moving a term without flipping its sign. Transposition is legal only with a sign change. From $5x=2x+12$, the slip writes $5x+2x=12$ and lands on a fraction. Correct: $5x-2x=12$, so $3x=12$. Ray's rule for every solve for the x: a term crosses the equals sign only with its sign changed.

3. Dropping a minus in front of a bracket. In $6x-(3x+8)=16$, the slip writes $6x-3x+8=16$. The minus hits the whole bracket: $6x-3x-8=16$, so $3x=24$ and $x=8$. Put a 1 in front of the bracket and distribute - the standard rescue before you solve for the x any further.

4. Dividing only one term. From $2x+4=10$, the slip divides only the $2x$ by 2 and writes $x+4=10$. The 4 and the 10 must ride along: $x+2=5$, then $x=3$. Divide whole sides, every term.

5. Stopping early or skipping the check. Two ways to stop too soon. One: leaving a coefficient behind, as at $P-6=2w$ - a solve for the x ends only at coefficient 1. Two: forgetting a second root when $x$ is squared - Example 8 owed two values. And never skip the verification: two lines of arithmetic, and you solve for the x with proof in hand.