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Equations with the variable on both sides

Collecting the letters on one side, and the two strange cases.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

In this lesson you solve equations where the letter appears on both sides, by collecting it on one. Two odd outcomes can arise and both mean something: if you end with $5 = 5$ the equation is true for every value, and if you end with $5 = 7$ it is true for none. Those are not mistakes. They are the answer, and recognizing them is part of the skill.

2. What you already know

You can solve an equation such as $5x - 8 = 27$ by undoing operations in reverse order, doing the same thing to both sides each time. You can expand a bracket, such as $3(x + 4) = 3x + 12$, and collect like terms, so $6x - 2x$ becomes $4x$. You know that a solution is a value that makes both sides equal, and that you can check it by substituting into the original equation. This lesson adds one new move: when $x$ appears on both sides, you first gather the $x$ terms together on one side. After that, the equation is the kind you already know how to solve.

3. Words this lesson uses

4. Gather the letters on one side

Look at $7x + 5 = 3x + 29$. The unknown appears on both sides, so you cannot just undo operations: there is no single chain of steps from $x$ to a number.

The fix is to make one side free of $x$. Think of the balance scale again. The left pan holds seven bags of marbles and $5$ loose marbles. The right pan holds three bags and $29$ loose marbles. Every bag holds the same unknown number. Take three bags off each pan, and the scale still balances:

$$7x + 5 - 3x = 3x + 29 - 3x \quad\Rightarrow\quad 4x + 5 = 29.$$

Now $x$ is on the left only, and you know what to do: $4x = 24$, so $x = 6$. Check in the original: $7 \times 6 + 5 = 47$ and $3 \times 6 + 29 = 47$.

Which side should the $x$ terms go to? Either works, but it is easiest to subtract the smaller $x$ term, so the $x$ term that is left has a positive coefficient. Here $3x$ is smaller than $7x$, so you subtract $3x$.

Sometimes the $x$ terms cancel completely. Then there is no $x$ left, just a statement about numbers. If that statement is true, like $5 = 5$, the equation is true for every value of $x$: it has infinitely many solutions. If it is false, like $5 = 9$, it is true for no value of $x$: it has no solution. Those are not mistakes. They are the answer.

5. Why a term seems to change sign

People often say "move $3x$ to the other side and change its sign". That shortcut works, but it hides what is really happening, and that is where mistakes come from.

Nothing moves. In $7x + 5 = 3x + 29$, you subtract $3x$ from both sides. On the right, $3x - 3x = 0$, so the $3x$ disappears. On the left, $7x - 3x = 4x$. The $3x$ did not travel; you took it away from both pans.

The same is true for a subtracted term. In $2x + 9 = 30 - 5x$, the right side has $-5x$. To clear it, add $5x$ to both sides: $-5x + 5x = 0$ on the right, and $2x + 5x = 7x$ on the left. The equation becomes $7x + 9 = 30$, so $7x = 21$ and $x = 3$.

If you always say "add ... to both sides" or "subtract ... from both sides", you will never forget the second side, and you will never flip a sign that should have stayed.

6. Collecting on the right is fine

Suppose the larger $x$ term is on the right, as in $2x - 9 = 5x + 12$. You could subtract $5x$ from both sides and get $-3x - 9 = 12$. That works, but now you must divide by $-3$ at the end, which is an easy place to lose a sign.

Instead, subtract the smaller term, $2x$, from both sides: $-9 = 3x + 12$. Now the $x$ term is on the right with a positive coefficient. Subtract $12$ from both sides: $-21 = 3x$. Divide by $3$: $-7 = x$.

An equation can be read in either direction, so $-7 = x$ says exactly the same thing as $x = -7$. Write the answer with $x$ first if you like, but there is no need to force $x$ onto the left side while you are solving.

Either route gives the same answer, because every move keeps the two sides equal. Choosing the route with positive coefficients is simply a way to make fewer slips.

7. Brackets, fractions and decimals on both sides

Clear the clutter before collecting. Brackets: expand them on each side and collect like terms on that side, so each side has at most one $x$ term and one number. For $3(x - 2) + x = 2(x + 5)$, the left becomes $4x - 6$ and the right becomes $2x + 10$. Then collect as usual: $2x = 16$, $x = 8$.

Fractions: multiply every term on both sides by a number that all the denominators divide into. For $\dfrac{x}{2} + 1 = \dfrac{x}{3} + 3$, multiply by $6$: $3x + 6 = 2x + 18$, so $x = 12$. Every term must be multiplied, including the whole numbers.

Decimals: multiply by $10$ or $100$ to make the coefficients whole. For $0.4x + 1.5 = 0.1x + 3$, multiply by $10$: $4x + 15 = x + 30$, so $3x = 15$ and $x = 5$. You can also solve with the decimals as they are, but whole numbers make the arithmetic easier to check.

8. One solution, no solution, or infinitely many

When you collect the $x$ terms, one of three things happens.

One solution. Some $x$ is left over, as in $4x = 24$. Divide and you get one value, $x = 6$. This happens whenever the two sides have different coefficients of $x$.

No solution. The $x$ terms cancel and a false statement is left. In $3x + 4 = 3x - 1$, subtract $3x$: $4 = -1$. That is never true, so no number works. This happens when the coefficients are the same but the constant terms are different.

Infinitely many solutions. The $x$ terms cancel and a true statement is left. In $2(x + 3) = 2x + 6$, expand: $2x + 6 = 2x + 6$. Subtract $2x$: $6 = 6$. That is always true, so every number is a solution. This happens when the two sides are the same expression once simplified.

You can often tell which case you have before solving. Simplify each side into the form $ax + b$. Compare the coefficients first, then the constants. If you want a check, try $x = 0$ and $x = 1$ in the original equation: for an identity both work, and for a no-solution equation neither does.

9. The method, step by step, and how to check it

1. Simplify each side on its own: expand brackets, clear fractions or decimals, and collect like terms.
2. Collect the $x$ terms on one side by adding or subtracting an $x$ term on both sides. Subtract the smaller one to keep the coefficient positive.
3. Look at what is left. If the $x$ terms cancelled, read the statement: true means infinitely many solutions, false means no solution. Stop there.
4. Otherwise, undo the number term on both sides, then divide by the coefficient of $x$.
5. Check by substituting into the original equation.

Why the moves are allowed. Adding or subtracting the same thing on both sides keeps them equal, whether that thing is a number or a term like $3x$. $3x$ stands for a number too, so the rule does not change.

How to check. Work out the left side and the right side separately with your value of $x$. With $x$ on both sides, both sides change as $x$ changes, so both must be worked out. They should give the same number. If they do not, look first at the step where you collected the $x$ terms.

10. In the world: the temperature that reads the same

Is there a temperature that is the same number in Fahrenheit and Celsius? The two scales are linked by $F = 1.8C + 32$. If the readings are equal, call both of them $t$: then $t = 1.8t + 32$. The variable is on both sides. Subtract $1.8t$ from both sides: $-0.8t = 32$. Divide by $-0.8$: $t = -40$. So $-40$ degrees Fahrenheit is exactly $-40$ degrees Celsius. Check: $1.8 \times (-40) + 32 = -72 + 32 = -40$. Places such as northern Minnesota and Alaska see temperatures this low in the coldest winters, and at that point it does not matter which scale the thermometer uses.

11. In the world: renting or buying

A family plans to go skiing several days each winter. Renting skis costs $40$ dollars a day. Buying skis costs $400$ dollars, plus about $8$ dollars a day for tuning and waxing. After how many days are the costs equal? Let $d$ be the number of days: $40d = 400 + 8d$. Subtract $8d$: $32d = 400$, so $d = 12.5$. Renting is cheaper for $12$ days or fewer, and buying is cheaper from $13$ days on. The equation gives the break-even point; the family decides by how many days they really expect to ski.

12. In the world: catching up

A school bus leaves at $8{:}00$ and drives at $30$ miles per hour. A student who missed it gets a ride $20$ minutes later, at $45$ miles per hour. When does the car catch the bus? Let $t$ be the hours since $8{:}00$. The bus has gone $30t$ miles and the car $45\left(t - \dfrac{1}{3}\right)$ miles. At the catch-up point they are equal: $30t = 45t - 15$. Subtract $30t$: $0 = 15t - 15$, so $t = 1$. The car catches the bus at $9{:}00$, $30$ miles from school.

13. Mistakes to watch for

Changing only one side. Subtracting $3x$ on the left but not on the right gives a new equation with a different solution.

Adding the $x$ terms instead of cancelling them. From $7x + 5 = 3x + 29$, the $x$ term becomes $4x$, not $10x$.

Calling $0 = 0$ or $5 = 5$ an error. A true statement with no $x$ means every value is a solution.

Writing $x = 0$ when the $x$ terms cancel. Cancelling means $x$ has gone, not that $x$ is zero.

Checking only one side. With $x$ on both sides, both sides depend on $x$; work out each one.

14. Subtracting the smaller $x$ term

1. Solve $7x + 5 = 3x + 29$. Subtract $3x$ from both sides.

   $7x - 3x + 5 = 3x - 3x + 29$

   $3x$ is the smaller $x$ term, so the $x$ term left behind is positive.

2. Simplify both sides.

   $4x + 5 = 29$

   $7x - 3x = 4x$, and $3x - 3x = 0$ on the right.

3. Subtract $5$ from both sides.

   $4x = 24$

   The same amount off both sides keeps them equal.

4. Divide both sides by $4$.

   $x = 6$

   Dividing undoes multiplying by $4$.

5. Check both sides of the original equation.

   $7(6) + 5 = 47, \quad 3(6) + 29 = 47$

   Both sides give $47$, so $x = 6$ is the solution.

15. Collecting on the right, with a negative answer

1. Solve $2x - 9 = 5x + 12$. Subtract $2x$ from both sides.

   $2x - 2x - 9 = 5x - 2x + 12$

   The smaller $x$ term is on the left, so the $x$ terms gather on the right.

2. Simplify both sides.

   $-9 = 3x + 12$

   $5x - 2x = 3x$.

3. Subtract $12$ from both sides.

   $-9 - 12 = 3x$

   Undo the number term on the side with $x$.

4. Simplify the left side.

   $-21 = 3x$

   $-9 - 12 = -21$.

5. Divide both sides by $3$.

   $-7 = x$

   A negative divided by a positive is negative.

6. Check the left side with $x = -7$.

   $2(-7) - 9 = -14 - 9 = -23$

   Brackets keep the sign with the number.

7. Check the right side.

   $5(-7) + 12 = -35 + 12 = -23$

   Both sides give $-23$, so $x = -7$.

16. Brackets on both sides

1. Solve $4(x - 3) + 2x = 2(x + 8) + 4$. Expand the bracket on the left.

   $4x - 12 + 2x = 2(x + 8) + 4$

   $4 \times x = 4x$ and $4 \times (-3) = -12$.

2. Collect like terms on the left.

   $6x - 12 = 2(x + 8) + 4$

   $4x + 2x = 6x$.

3. Expand the bracket on the right.

   $6x - 12 = 2x + 16 + 4$

   $2 \times x = 2x$ and $2 \times 8 = 16$.

4. Collect like terms on the right.

   $6x - 12 = 2x + 20$

   $16 + 4 = 20$. Each side now has one $x$ term and one number.

5. Subtract $2x$ from both sides.

   $4x - 12 = 20$

   $6x - 2x = 4x$, and the right side has no $x$ left.

6. Add $12$ to both sides.

   $4x = 32$

   Adding $12$ undoes subtracting $12$.

7. Divide both sides by $4$.

   $x = 8$

   Dividing undoes multiplying by $4$.

8. Check the left side with $x = 8$.

   $4(8 - 3) + 2(8) = 20 + 16 = 36$

   Work inside the bracket first.

9. Check the right side.

   $2(8 + 8) + 4 = 32 + 4 = 36$

   Both sides give $36$, so $x = 8$ is the solution.

17. Your turn: solve $2(3x - 1) = 6x + 7$.

1. Expand the bracket on the left.

   $6x - 2 = 6x + 7$

   $2 \times 3x = 6x$ and $2 \times (-1) = -2$.

2. Subtract $6x$ from both sides.

   $-2 = 7$

   The $x$ terms are the same, so they cancel.

3. Your turn: work this step out. Its working is at the end of the packet.

   Read what is left.

4. Your turn: work this step out. Its working is at the end of the packet.

   State the answer.

18. Guided practice

How many solutions does $2(x - 1) = 2x - 8$ have?

• No solution
• Exactly one solution
• Infinitely many solutions
• One solution, $x = 0$

19. Guided practice

Complete the worked solution of $3x + 23 = 8x - 7$.

1. Subtract $3x$ from both sides.

   $23 =$ _p $x - 7$

   The right side has more $x$, so collecting there keeps the coefficient positive.

2. Add $7$ to both sides and work out the left side.

   $23 + 7 =$ _q

   Adding $7$ cancels the $-7$ on the right, leaving only the $x$ term there.

3. Divide both sides by the coefficient of $x$.

   $x =$ _s

   Dividing undoes the multiplying, and $x$ may sit on the right side of the equals sign.

20. Guided practice

Solve $6x - 4 = 6 - 4x$. Write one equation per line, ending with $x = $ your answer.

6x - 4 = 6 - 4x

21. Practice

Solve $6(x - 2) = 3(x - 1) - 24$. Write one equation per line, ending with $x = $ your answer.

6(x - 2) = 3(x - 1) - 24

22. Practice

How many solutions does $2x - 8 + 2x = 4x - 8$ have?

• Infinitely many solutions
• Exactly one solution
• No solution
• One solution, $x = -8$

23. Practice

Solve $\dfrac{x}{3} + 2 = \dfrac{x}{6} + 5$. Write one equation per line, ending with $x = $ your answer.

x/3 + 2 = x/6 + 5

24. Somewhere new

Riverside gym charges a joining fee of $28$ dollars and $25$ dollars a month. Hillside gym charges a joining fee of $64$ dollars and $21$ dollars a month. After how many months will you have paid the same at both, and how much will that be?

After _months months, both gyms will have cost _cost dollars.

25. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

26. Test question

Solve $0.5x + 4 = 0.1x + 8$. Write one equation per line, ending with $x = $ your answer.

0.5x + 4 = 0.1x + 8

27. What you can do now

You can solve an equation with the variable on both sides. Without looking: what does it mean if you end up with $3 = 3$, and what if you end up with $3 = 8$?

Working for the steps left to you

17. Your turn: solve $2(3x - 1) = 6x + 7$., step 3

17. Your turn: solve $2(3x - 1) = 6x + 7$., step 4