Linear Equations: Variables on Both Sides, Parentheses and Fractions

An expression gives a calculation, such as $3x+5$. An equation says two expressions are equal, such as $3x+5=2x+9$. The equals sign means “has the same value as.” It does not mean “the answer comes next.”

What the letter means

A variable, also called an unknown, stands for a number. Each occurrence of $x$ in an equation stands for that same number. In $3x+5=2x+9$, you cannot choose one value for the left-hand $x$ and a different value for the right-hand $x$.

A solution makes both sides equal. To solve an equation, find all its solutions. To simplify an expression, write the same calculation in a simpler way; there is no equality to solve.

Test a value on both sides

Does $x=4$ solve $3x+5=2x+9$? Substitute: replace each $x$ with $4$. Calculate the sides separately.

• Left side: $3(4)+5=12+5=17$.
• Right side: $2(4)+9=8+9=17$.

Both sides give $17$, so $x=4$ is a solution. The solution is $4$, the number replacing the variable. The shared value $17$ is what each side gives.

Now test $x=3$. The left side gives $3(3)+5=14$. The right side gives $2(3)+9=15$. These values differ, so $3$ fails. This tells us about the number we tried; the equation still has the solution $4$.

Check the original equation

After solving, substitute into the equation you started with. A later line might already contain a mistake. Checking only that line can hide the mistake.

Suppose someone changes $2x+6=10$ into $x+6=10$, then gets $x=4$. Their final line agrees with their answer. But the original left side gives $2(4)+6=14$, not $10$. The check catches the earlier error.

Keep a negative value together

Use parentheses when replacing a variable with a negative number. For $x=-2$ in $5-3x$, write:

The $3$ still multiplies the substituted number. Writing $5-3-2$ loses that multiplication.

Exact answers can take different forms

A solution can be negative or fractional. For $2x+1=6$, the solution is $x=\frac{5}{2}=2.5$. For $x+9=4$, it is $x=-5$. Keep fractions exact unless rounding is requested.

These are linear equations in one variable. Their sides contain numbers and number multiples of the variable, including grouped terms such as $2(x+5)$ or $\frac{x}{3}$. The variable is not squared or multiplied by itself, and it is not in a denominator.

Think of an equation as a balanced scale. Both pans hold the same total. Removing the same amount from each pan keeps the scale balanced. An equation works the same way, even when its terms include negative numbers.

Do the same thing to both sides

The properties of equality allow you to:

• Add the same number or term to both sides.
• Subtract the same number or term from both sides.
• Multiply both sides by the same nonzero number.
• Divide both sides by the same nonzero number.

The new equation has exactly the same solutions. Such equations are called equivalent equations. Each step must act on both complete sides.

Undo operations

Inverse operations undo each other. Addition undoes subtraction; division undoes multiplication. Solve $4x+7=31$ by removing the added number first.

1. Subtract $7$ from both sides: $4x+7-7=31-7$, so $4x=24$.
2. Divide both sides by $4$: $x=6$.
3. Check the start: $4(6)+7=31$.

The reason for the first step is subtraction on both sides. The reason for the second is division on both sides. These reasons explain why the answer still belongs to the original equation.

Simplifying is also allowed

Combining like terms keeps a side's value unchanged. In $3x+2x-4=21$, combine $3x+2x$ to get $5x-4=21$. Add $4$ on both sides: $5x=25$. Divide by $5$: $x=5$.

You can choose a different order. Dividing $5x-4=21$ by $5$ first gives $x-\frac{4}{5}=\frac{21}{5}$. Adding $\frac{4}{5}$ then gives $x=5$. Adding $4$ first is easier here because it avoids fractions during the work.

A whole side means every term

For $2x+6=10$, dividing by $2$ gives $x+3=5$. Both terms on the left were divided, as was the right side. Dividing only $2x$ would give $x+6=10$, which changes the solution.

You can instead subtract $6$ first. This gives $2x=4$, then $x=2$. A clear route is useful, but the balance rule decides whether a route is valid.

Why zero is different

Division by zero is undefined. Multiplying by zero also loses information needed for solving: $x=3$ would become $0=0$. Every number satisfies the new statement, although only $3$ satisfies the start. That step does not give an equivalent equation.

The shortcut “move a term and change its sign” records addition or subtraction on both sides. It does not apply to a multiplier. From $3x=12$, divide by $3$ to get $x=4$. Subtracting $3$ from $12$ does not undo multiplication.

Algebra gives exact answers, but it is not the only way to find a solution. A chain of operations can be undone. A table or a graph can show where two sides become equal.

Work backward through a chain

“I multiply a number by $3$, subtract $5$, and get $16$.” The equation is $3x-5=16$. Undo the last operation first: $16+5=21$. Then undo multiplication: $21÷3=7$.

The number is $7$. Check in the forward direction: $3(7)-5=16$. Working backward is helpful when the unknown appears once in one chain. When it appears on both sides, balance steps collect its terms together.

Try, check, and improve

Trial and improvement means choosing a value, calculating both sides, then using the result to choose a better value. For $3x+7=5x-4$, start with two whole numbers.

At $5$, the left side is larger. At $6$, the right side is larger. The solution lies between them. Trying the halfway value finds an exact match, so $x=5.5$.

The table organizes the trials. Both sides must be calculated for the same input on each row. A row with nearly equal outputs suggests a nearby solution; equal outputs give a checked solution.

Algebra confirms the match: subtract $3x$ from both sides, giving $7=2x-4$. Add $4$: $11=2x$. Divide by $2$: $x=5.5$.

Read the horizontal coordinate

To solve $2x+1=7-x$ graphically, compare the lines $y=2x+1$ and $y=7-x$ on the same axes. They cross at $(2,5)$.

The horizontal coordinate gives $x=2$. The vertical coordinate, $5$, is the value both sides share. Substitution confirms this: $2(2)+1=5$ and $7-2=5$.

A graph reading is usually an estimate. If the crossing appears near $x=1.7$, write $x\approx 1.7$. The symbol $\approx$ means “approximately equal to.” Use algebra or an exact substitution check before calling a graph reading exact.

Choose a method that fits

Working backward is quick for a single chain. Tables help you see the gap between two amounts. Graphs show where the amounts meet. Trial and improvement may need many attempts when the solution is a fraction such as $\frac{1}{3}$. Balance steps can find that exact fraction directly.

In $5x+3=2x+18$, both sides depend on $x$. The aim is to collect the variable terms on one side and the numbers on the other. Subtracting the same variable term from both sides keeps the equation balanced.

Remove one variable term

Subtract $2x$ from both sides:

Combine like terms. The right-hand variable terms cancel, leaving:

This is now a two-step equation. Subtract $3$ from both sides to get $3x=15$. Divide by $3$ to get $x=5$.

Check each original side: $5(5)+3=28$ and $2(5)+18=28$. On a balance picture, removing $2x$ from each pan removes equal amounts. That is why the first step keeps the solution.

Either side can hold the variable

You could subtract $5x$ instead. Then $3=-3x+18$. Subtract $18$ to get $-15=-3x$, and divide by $-3$ to get $5=x$.

The answer is unchanged. Removing the smaller positive variable term was easier because it left a positive coefficient. A coefficient is the number multiplying the variable; the coefficient of $3x$ is $3$.

An equation reads in either direction. The statements $15=3x$ and $3x=15$ mean the same thing. There is no need to rearrange a correct line just to put the variable on the left.

Follow subtraction with its inverse

Solve $7x-4=3x+10$.

1. Subtract $3x$ from both sides: $4x-4=10$.
2. Add $4$ to both sides: $4x=14$.
3. Divide by $4$: $x=\frac{14}{4}=3.5$.

The $-4$ needs addition to remove it. Subtracting another $4$ would leave $-8$. Checking the original sides gives $7(3.5)-4=20.5$ and $3(3.5)+10=20.5$.

A lone variable still counts

The term $x$ means $1x$. For $x+12=4x$, subtract $x$ from both sides. This gives $12=3x$, so $x=4$. The original sides both give $16$.

Combining $x$ with $4x$ across the equals sign without a balance step would be an error. The sign depends on the operation you apply. From $5x+3=2x+18$, subtracting $2x$ gives $3x+3=18$; it does not give $7x+3=18$.

A sign belongs to its term. In $7-2x=3x-8$, the variable terms are $-2x$ and $3x$. Keeping those signs attached helps you choose a useful balance step.

Add a negative term away

Add $2x$ to both sides of $7-2x=3x-8$. This gives $7=5x-8$. Add $8$: $15=5x$. Divide by $5$: $x=3$. Both original sides give $1$.

For $-4x+9=2x-3$, adding $4x$ gives $9=6x-3$. Adding $3$ then gives $12=6x$, so $x=2$. Both sides give $1$ again, but the solution is the input $2$.

A negative solution is a valid answer

Solve $2x+11=5x+20$. Subtract $2x$: $11=3x+20$. Subtract $20$: $-9=3x$. Divide by $3$: $x=-3$.

Check with parentheses: $2(-3)+11=5$ and $5(-3)+20=5$. A negative answer works because it makes the original sides equal.

When dividing, keep the coefficient's sign. From $-3x=12$, divide by $-3$, giving $x=-4$. Using positive $3$ would give the wrong sign. Also remember that subtracting a negative adds its opposite: $5-(-3)=8$.

Keep fractions exact

For $6x-1=2x+5$, subtract $2x$ and add $1$. This leaves $4x=6$, so $x=\frac{6}{4}=\frac{3}{2}=1.5$.

For $5x+2=2x+3$, subtraction gives $3x=1$. The solution is exactly $\frac{1}{3}$. The decimal $0.33$ stops too soon: it makes the left side $3.65$ and the right side $3.66$. Use the fraction when the decimal does not terminate.

Clear decimals across the whole equation

Solve $0.4x+1.2=0.1x+3$. Multiplying both sides by $10$ makes every coefficient and constant a whole number:

Subtract $x$ to get $3x+12=30$. Subtract $12$: $3x=18$. Divide by $3$: $x=6$. The check gives $0.4(6)+1.2=3.6$ and $0.1(6)+3=3.6$.

The whole-number term $3$ was multiplied too. Leaving it unchanged would produce a different equation. With hundredths, multiplying every term by $100$ may be useful. You can also solve with the decimals directly; clearing them is a way to simplify the arithmetic.

The distributive property connects multiplication with addition: $a(b+c)=ab+ac$. To expand a group, multiply each inside term by the outside number. Thus $3(x+4)=3x+12$.

Expand, then solve

For $3(x+4)=27$, expand the left side:

Subtract $12$ from both sides: $3x=15$. Divide by $3$: $x=5$. Check the original group: $3(5+4)=3(9)=27$.

Writing $3x+4$ would leave the second term unmultiplied. One way to see the error is to try $x=1$. Three groups of $1+4$ total $15$, while $3(1)+4$ gives only $7$.

Divide first when the whole side is a product

In $3(x+4)=27$, the left side is three copies of one group. Divide both sides by $3$ first: $x+4=9$. Then subtract $4$ to get $x=5$.

Both routes keep the equation balanced. Dividing first avoids expansion here. But in $5+2(x+3)=21$, the $5$ is outside the product. Expand to get $5+2x+6=21$. Combine the constants: $2x+11=21$, so $2x=10$ and $x=5$. Adding $5+2$ first would wrongly turn an addition into part of the multiplier.

Multiply by the negative sign too

Expand $-2(x-5)$ by multiplying $x$ and $-5$ by $-2$:

The second term is positive because a negative times a negative is positive. For $-2(x-5)=4$, this gives $-2x+10=4$. Subtract $10$: $-2x=-6$. Divide by $-2$: $x=3$. Check: $-2(3-5)=-2(-2)=4$.

A variable can appear outside the group

Solve $2(3x-1)=4x+9$. Expanding gives $6x-2=4x+9$. Subtract $4x$, then add $2$: $2x=11$. Therefore $x=5.5$. The original sides both give $31$.

A fraction multiplier works the same way. For $\frac{1}{2}(4x+6)=x+9$, expansion gives $2x+3=x+9$. Subtract $x$ and $3$ to get $x=6$. Checking gives $\frac{1}{2}(30)=15$ and $6+9=15$.

With several groups, expand each one before collecting like terms. Keep an equals sign on every line. This makes it possible to follow what happens on each side.

Expand both sides

Solve $4(2x-3)=5(x+3)$.

1. Expand both groups: $8x-12=5x+15$.
2. Subtract $5x$ from both sides: $3x-12=15$.
3. Add $12$: $3x=27$.
4. Divide by $3$: $x=9$.

In the original equation, $4(18-3)=60$ and $5(9+3)=60$. Each factor multiplied both terms in its own group.

Subtracting a group changes every sign

A minus sign before parentheses means multiplication by $-1$. So $-(4-x)=-4+x$: the $4$ becomes $-4$, and the $-x$ becomes $+x$.

For $x-(4-x)=10$, expansion gives $x-4+x=10$. Combine the variable terms: $2x-4=10$. Then $2x=14$ and $x=7$. Check: $7-(4-7)=7-(-3)=10$.

Keep a minus with its multiplier

Solve $5(x-1)-2(x-3)=13$. The second group has multiplier $-2$, not $2$.

Collect like terms: $3x+1=13$. Subtract $1$: $3x=12$. Divide by $3$: $x=4$. The original left side is $5(3)-2(1)=13$.

The tempting expansion $-2x-6$ has the wrong constant sign. The product $(-2)(-3)$ is $+6$. Writing the signed multiplier beside each product helps keep this visible.

Groups can have extra terms around them

For $3(2x+1)-4=2(x+5)+1$, simplify each side separately:

• Left side: $6x+3-4=6x-1$.
• Right side: $2x+10+1=2x+11$.

Now solve $6x-1=2x+11$. Subtract $2x$ and add $1$: $4x=12$, so $x=3$. Both original sides give $17$.

Decimal factors also distribute. The equation $0.5(x+6)=0.2(x+30)$ becomes $0.5x+3=0.2x+6$. Thus $0.3x=3$ and $x=10$. Both sides give $8$.

Write different stages on different lines. A chain such as $8x-12=5x+15=3x-12$ claims all three expressions are equal. The last expression came from a new balance step, so it belongs in a new equation.

The expression $\frac{x}{4}$ means $x$ divided by $4$. It is also $\frac{1}{4}x$. Multiplication undoes this division, so multiplying both sides by $4$ removes the denominator.

Remove an outside addition first

Solve $\frac{x}{4}+3=7$. Subtract $3$ from both sides:

Multiply both sides by $4$: $x=16$. Check: $\frac{16}{4}+3=4+3=7$.

You could multiply the original equation by $4$ first. That gives $x+12=28$, then $x=16$. The $3$ must become $12$ because the multiplier acts on the whole side.

Undo a fraction coefficient

A reciprocal swaps a fraction's numerator and denominator. The reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$. Their product is $1$, so multiplying $\frac{2}{3}x=8$ by $\frac{3}{2}$ gives $x=12$.

Another route is to multiply by $3$, giving $2x=24$, then divide by $2$. Both routes undo the same coefficient. The check is $\frac{2}{3}(12)=8$.

The bar groups the numerator

Compare $\frac{x+5}{3}$ with $\frac{x}{3}+5$. In the first expression, both $x$ and $5$ are divided by $3$. In the second, $5$ is added after the division.

For $\frac{x+5}{3}=4$, multiply both sides by $3$: $x+5=12$. Subtract $5$: $x=7$. The check gives $\frac{7+5}{3}=4$.

Subtracting $5$ from just the numerator and the right side would give the false step $\frac{x}{3}=-1$. The numerator's $5$ contributes $\frac{5}{3}$ to the left side, not $5$.

Multiply every term on the other side

For $\frac{2x-1}{5}=x-2$, multiply both complete sides by $5$:

Expand to get $2x-1=5x-10$. Add $1$ and subtract $2x$: $9=3x$, so $x=3$. Both original sides give $1$. Parentheses keep the right-hand subtraction inside the multiplication.

Keep signs when a fraction is negative. For $-\frac{x}{3}=5$, multiply both sides by $-3$ to get $x=-15$. Checking gives $-\frac{-15}{3}=5$.

To clear several fractions, choose a number divisible by every denominator. Multiplying the whole equation by that number removes all the denominators at once. A least common multiple keeps the new numbers as small as possible.

Choose one multiplier for the whole equation

For $\frac{x}{3}+\frac{x}{4}=7$, the denominators are $3$ and $4$. Their least common multiple is $12$. Multiply every term by $12$:

Combine like terms: $7x=84$, so $x=12$. The original check is $\frac{12}{3}+\frac{12}{4}=4+3=7$.

The multiplier for each numerator comes from dividing $12$ by its denominator. Thus $12÷3=4$ and $12÷4=3$. The right-hand $7$ becomes $84$.

Whole numbers are terms too

For $\frac{x}{2}+1=\frac{x}{3}+4$, use $6$:

Subtract $2x$ and $6$: $x=18$. Both original sides give $10$. Leaving $1$ and $4$ unchanged would multiply only part of each side, changing the equation.

Keep each numerator together

Solve $\frac{x+1}{2}-\frac{x-2}{3}=2$. Multiplying by $6$ gives:

The minus applies to the whole second numerator. Expand: $3x+3-2x+4=12$. Combine terms: $x+7=12$, so $x=5$. Check: $\frac{6}{2}-\frac{3}{3}=3-1=2$.

The constant from $-2(x-2)$ is $+4$. Writing $-2x-4$ loses the negative-times-negative rule. Writing $-2x-2$ fails to multiply the whole numerator.

A longer fractional equation

For $\frac{4x+3}{10}+\frac{6x-5}{5}=\frac{13}{2}$, use $10$. The three numerator multipliers are $1$, $2$, and $5$:

Expand and collect: $16x-7=65$. Add $7$: $16x=72$. Divide by $16$: $x=4.5$. Checking gives $\frac{21}{10}+\frac{22}{5}=2.1+4.4=6.5=\frac{13}{2}$.

Any common multiple works. Multiplying the first example by $24$ gives $8x+6x=168$, which still gives $x=12$.

A linear equation in one variable has one solution, no solution, or infinitely many solutions. Use valid balance steps until the remaining statement makes the case clear.

A number left for the variable

Solve $4x+1=2x+1$. Subtract $2x$ from both sides, then subtract $1$. This gives $2x=0$. Dividing by $2$ gives $x=0$.

There is one solution, zero. Substitution confirms it: both original sides give $1$. Zero is a number that can solve an equation; it does not mean “no solution.”

A false statement means no solution

For $3x+5=3x-1$, subtract $3x$ from both sides. You get $5=-1$. This is false, so there is no solution.

No choice of $x$ can remove the gap. Both sides contain the same $3x$, but the left side is always $6$ greater. The solution set, meaning all the solutions together, is empty.

There is no division step left to do. In particular, a vanished variable does not become $x=0$. Read what remains after cancellation.

A true statement means infinitely many solutions

Expand $2(x+3)=2x+6$. The equation becomes $2x+6=2x+6$. Subtract $2x$ to get $6=6$.

This statement is true for every value of $x$. The original sides are two forms of the same expression. Every number is a solution, so there are infinitely many solutions. An equation true for every value is an identity.

For example, $x=0$ gives $6=6$, and $x=10$ gives $26=26$. These tests show how the identity works. Expanding proves it for all values at once.

Simplify before deciding

Consider $5x-2(x+1)=3x-2$. Expansion gives $5x-2x-2=3x-2$, or $3x-2=3x-2$. It is an identity.

Now change the last constant: $5x-2(x+1)=3x+2$. Expansion and subtraction leave $-2=2$, a false statement. This equation has no solution.

The two constant statements may use other numbers. What matters is whether the final variable-free equality is true or false.

After expanding and collecting like terms, each side has a variable term and a constant. Compare the variable coefficients first. If they match, compare the constants too.

Compare simplified sides

For an equation written as $ax+b=cx+d$, the letters $a$ and $c$ are fixed coefficients. The letters $b$ and $d$ are fixed constants.

For $6x-4=6x+1$, the coefficients match but the constants differ. There is no solution. For $6x-4=2(3x-2)$, expand the right side to $6x-4$. Both sides match, so every number works.

For $6x-4=5x+1$, the coefficients differ. Subtraction gives $x=5$, the one solution. Equal constants alone do not make an identity: $4x+1=2x+1$ has different coefficients and the single solution $x=0$.

Build an equation with a chosen result

Expand the left side of $4(x-3)=4x+\mathrm{\square }$. It is $4x-12$. Filling the box with $-12$ makes an identity. Any other constant leaves equal coefficients but different constants, so there is no solution.

To get one solution, change a coefficient instead. The equation $4(x-3)=3x+1$ becomes $4x-12=3x+1$, giving $x=13$.

Give the result a meaning

Gym A costs $30 to join plus $15 per month. Gym B costs $10 to join plus $15 per month. Equality would require $30+15m=10+15m$. This reduces to $30=10$. The costs never match: Gym A keeps its $20 starting gap.

If Gym B instead costs $20 per month, solve $30+15m=10+20m$. This gives $20=5m$, so $m=4$. Both totals are $90. Different rates give one algebraic meeting value. Check that the value fits the situation; a negative month would not be a meeting after joining.

An identity can describe equal counts in different forms. Three boxes with $n+2$ pens each contain $3(n+2)=3n+6$ pens. That equality cannot determine $n$: it works for every number, and for every allowed whole-number count.

On a graph, one solution is one crossing. No solution gives distinct parallel lines. Infinitely many solutions give the same line drawn twice.

Define the variable before writing the equation. For example, “Let $c$ be the cost of one book in dollars.” The variable represents a number, not the object itself. “Let $n$ be the number of tickets” is clearer than “$n$ is tickets.”

Read the order and the group

“Three less than a number” means $n-3$. The number comes first, then three is taken away. The expression $3-n$ means something different.

“Twice the sum of a number and five” means $2(n+5)$. The sum is made first, and the whole sum is doubled. “Twice a number, plus five” means $2n+5$. Parentheses show which calculation the doubling acts on.

Words such as “more than” suggest addition; “times” suggests multiplication. The full sentence decides how the quantities fit. “Is” or “gives the same result” often marks an equality.

Match two descriptions of one result

“I multiply a number by $4$ and subtract $7$. I get the same result if I double the number and add $9$.” Let $n$ be the number. The two results are $4n-7$ and $2n+9$:

Subtract $2n$ and add $7$ to get $2n=16$. Thus $n=8$. The check gives $4(8)-7=25$ and $2(8)+9=25$.

Describe related quantities with one variable

Ella has three times as many stickers as Sam. Together they have $84$. Let $s$ be Sam's number of stickers. Ella has $3s$, so $s+3s=84$. This gives $4s=84$ and $s=21$. Ella has $63$ stickers.

The check uses both facts: $21+63=84$ and $63=3(21)$. If the question asks for Ella's count, stopping at $s=21$ leaves the requested quantity unfinished.

Consecutive numbers have a fixed gap

Consecutive integers follow one another: $n$, $n+1$, $n+2$. If three consecutive integers total $54$, write:

Collect terms: $3n+3=54$. Then $3n=51$ and $n=17$. The integers are $17$, $18$, and $19$. Consecutive even or odd integers rise by two, so their expressions are $n$, $n+2$, $n+4$.

Check a comparison with easy numbers

“There are four times as many students as teachers” means $s=4t$, where $s$ and $t$ count students and teachers. With one teacher, there should be four students. The reversed equation $4s=t$ fails this easy check.

A geometry fact links the expressions in a diagram or description. A perimeter adds all the side lengths. Equal sides have equal lengths. Known angle totals give another kind of equality.

Add every side of a rectangle

A rectangle has length $(2x+3)$ cm, width $x$ cm, and perimeter $36$ cm. There are two lengths and two widths:

Expand: $4x+6+2x=36$. Collect terms: $6x+6=36$. Subtract $6$ and divide by $6$ to get $x=5$.

The width is $5$ cm. The length is $2(5)+3=13$ cm. Check the perimeter: $2(13+5)=36$ cm. Adding one length and one width would give half the perimeter.

Use an angle total

A triangle's angles are $x^{\circ }$, $2x^{\circ }$, and $(3x+30{)}^{\circ }$. The angles total ${180}^{\circ }$:

This gives $6x=150$ and $x=25$. The angles are ${25}^{\circ }$, ${50}^{\circ }$, and ${105}^{\circ }$. Their sum is ${180}^{\circ }$.

Adjacent angles forming a straight angle also total ${180}^{\circ }$. If they are $(4x+10{)}^{\circ }$ and $(2x-10{)}^{\circ }$, write $4x+10+2x-10=180$. Thus $x=30$, giving angles of ${130}^{\circ }$ and ${50}^{\circ }$. Angles around a point total ${360}^{\circ }$.

Equal sides give equal expressions

The equal sides of an isosceles triangle measure $(3x-2)$ cm and $(x+8)$ cm. Set the lengths equal:

Subtract $x$ and add $2$: $2x=10$, so $x=5$. Each equal side is $13$ cm. The label $x$ alone is not a side length here; substitute it into the expressions.

Compare two whole perimeters

A square has side $(x+2)$ cm. An equilateral triangle has side $(x+5)$ cm. Their perimeters match, so:

Expand: $4x+8=3x+15$. Subtract $3x$ and $8$: $x=7$. The square's side is $9$ cm and the triangle's side is $12$ cm. Both perimeters are $36$ cm.

Check the requested measurements after solving. Side lengths and the angles in these shapes must be positive. A zero or negative result calls for a check of the equation and arithmetic before accepting the shape.

To find when two amounts are equal, describe each amount at the same time or input. Use the same variable for both. Then set the complete expressions equal.

A starting amount plus repeated growth

Ana has $50 and saves $8 per week. Ben has $90 and saves $3 per week. Let $w$ be the number of weeks. Their totals are $50+8w$ and $90+3w$ dollars.

Subtract $3w$ and $50$: $5w=40$, so $w=8$. After eight weeks, each has $114.

The gap gives a second check. Ben starts $40 ahead. Ana adds $5 more each week, so the gap closes in $40÷5=8$ weeks. Using the sum of their saving rates would count Ben's growth in the wrong direction.

Compare prices using the same usage

Plan A costs $20 per month plus $0.10 per call minute. Plan B costs $5 per month plus $0.25 per minute. Let $m$ count call minutes in that month:

Subtract $0.10m$ and $5$: $15=0.15m$. Thus $m=100$. At 100 minutes each plan costs $30. Below 100 minutes Plan B costs less; above 100 minutes Plan A costs less. The lower starting fee matters more for small usage, while the lower per-minute rate matters more for large usage.

Future ages change for both people

Maya is three times her brother's age. In six years she will be twice his age. Let $b$ be his age now; Maya's is $3b$. In six years their ages are $b+6$ and $3b+6$:

Expand: $3b+6=2b+12$. Subtract $2b$ and $6$: $b=6$. Their current ages are $6$ and $18$; their future ages are $12$ and $24$.

The multiplier acts on the brother's whole future age. Writing $2b+6$ would double only his current age before adding the years.

One amount can fall as the other rises

Tank A starts with $200$ liters and loses $15$ liters per minute. Tank B starts with $50$ liters and gains $10$ liters per minute. Let $t$ be minutes:

Add $15t$ and subtract $50$: $150=25t$, so $t=6$. Both tanks then hold $110$ liters. Here the gap closes by $25$ liters per minute because one level falls while the other rises.

Equal rates behave differently. With different starting amounts, the gap stays fixed and the amounts never match. With the same starting amount too, they match throughout.

An answer needs to pass two checks: it must solve the equation, and it must fit the original situation. A correctly solved equation can still describe the wrong relationship.

Check the words as well as the arithmetic

In the sticker problem, Sam has $21$ stickers and Ella has $63$. Their total is $84$, and Ella's count is three times Sam's. Both facts hold. If the question asks about Ella, the final sentence is “Ella has 63 stickers.” Reporting only Sam's variable value leaves the question unanswered.

Give units when they matter. “After eight weeks, Ana and Ben each have $114” names both the time and the amount. A bare number such as $8$ leaves the reader to guess its meaning.

Ask what kind of number is possible

People, tickets, and whole objects require nonnegative whole-number counts. A length must be positive. A length of $7.5$ cm is possible, and a temperature of $-4^{\circ }$ can be possible. Whether a negative or fractional value fits depends on what the variable measures.

Tickets cost $7 each, with one $12 booking fee. A bill of exactly $100 would require $7t+12=100$. Subtract $12$: $7t=88$, so $t=\frac{88}{7}\approx 12.57$.

That is not a whole number of tickets. Rounding to $13$ gives a bill of $103; using $12$ gives $96. Neither matches $100. The stated purchase has no whole-ticket answer.

Match the units before solving

A plumber charges $45 for a visit and $60 per hour. For a $135 bill, let $t$ be hours:

Subtract $45$ and divide by $60$: $t=1.5$ hours. That is $90$ minutes. The coefficient $60$ is a rate per hour, so putting a number of minutes beside it would describe the wrong cost. Convert units first or define the variable in the rate's unit.

Substitute into a formula

The formula $F=\frac{9}{5}C+32$ relates Fahrenheit and Celsius temperatures. For $F=77$, substitute the known value:

Subtract $32$: $45=\frac{9}{5}C$. Multiply by $\frac{5}{9}$: $C=25$. So ${77}^{\circ }$F equals ${25}^{\circ }$C. Substitution back gives $\frac{9}{5}(25)+32=77$.

A result can also describe a whole situation without giving one value. “The two plans never cost the same” explains no solution. “Both expressions count the same pens for every allowed box count” explains an identity. Neither statement means the unknown is zero.