Solve for X Calculator

An equation is a balance. Whatever you do to one side you must do to the other, and the balance still holds. Solving for $x$ means peeling operations off $x$ — using the inverse of each one — until $x$ stands alone on one side.

The inverses are the obvious pairs: addition undoes subtraction, multiplication undoes division, and squaring undoes a square root.

In $3x + 7 = 22$, two things have been done to $x$: it was multiplied by 3, then 7 was added. Undo them in reverse order, like taking off shoes after socks — subtract the 7 first, then divide by the 3:

$3x + 7 = 22 \;\Rightarrow\; 3x = 15 \;\Rightarrow\; x = 5$

That reverse order is the whole idea behind "two-step" equations, and it does not change no matter how many steps get added.

Before undoing anything, clean up each side:

1. Distribute through any brackets: $4(x + 3) = 4x + 12$. Multiply the outside term by every term inside.
2. Combine like terms on each side separately: $4x + 12 - 2x$ becomes $2x + 12$. Only terms with the identical variable part combine — $2x$ and $2x^2$ do not.
3. Clear fractions by multiplying every term by the least common denominator. $\frac{x+2}{3} = 4$ becomes $x + 2 = 12$ in one move.
4. Gather the variables on one side by adding or subtracting a variable term from both sides. Move them to whichever side keeps the coefficient positive — it prevents sign errors.
5. Undo the remaining two operations in reverse order.
6. Substitute your answer back into the original equation. This catches nearly every slip, and it costs ten seconds.

• Distributing to only the first term. $4(x + 3)$ is $4x + 12$, never $4x + 3$.
• Dropping a negative through brackets. $-2(x - 5) = -2x + 10$; the minus hits both terms.
• Operating on one side only. Subtracting 7 from the left means subtracting 7 from the right in the same line.
• Combining unlike terms. $3x + 5$ is already finished; it does not become $8x$.
• Dividing only part of a side. From $2x + 6 = 10$, dividing by 2 gives $x + 3 = 5$, not $x + 6 = 5$.
• Panicking when $x$ disappears. If you end at $0 = 8$ there is no solution; if you end at $0 = 0$ every value of $x$ works. Both are legitimate answers, not errors.