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Equations, inequalities and two variables

Test solutions by substituting, and solve one-step equations by undoing the operation on both sides.

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 learn what it means for a number to solve an equation and test values by substituting them. You solve the four kinds of one-step equation, $x + p = q$, $x - p = q$, $px = q$ and $x \div p = q$, with whole numbers, decimals and fractions, by doing the inverse operation to both sides. You write equations from stories and check every answer in the original equation.

2. What you already know

You know that addition and subtraction undo each other, and so do multiplication and division: $15 + 4 - 4 = 15$ and $15 \times 4 \div 4 = 15$. Earlier in this course you wrote expressions with letters, such as $n + 6$ or $3w$, and evaluated them by substituting a number for the letter. You can add, subtract, multiply and divide decimals, and you can divide by a fraction by multiplying by its reciprocal. This lesson puts those pieces together to answer a new kind of question: which number makes a statement true?

3. Words this lesson uses

4. Keep the balance, undo the operation

An equation says that two amounts are equal. $x + 4 = 11$ says that some number, plus 4, is 11. A solution is a number that makes the statement true. Try 7: $7 + 4 = 11$, true. Try 6: $6 + 4 = 10$, false. So 7 is the solution and 6 is not.

Guessing works for easy equations, but not for $x + 38.75 = 102.5$. Instead we undo what was done to the variable. In $x + 4 = 11$, 4 was added to $x$. The inverse of adding 4 is subtracting 4, so subtract 4:

$$x + 4 - 4 = 11 - 4, \qquad x = 7.$$

The key rule: whatever you do to one side, do to the other side. Picture a balance scale with $x + 4$ on the left pan and 11 on the right. It is level. Take 4 off the left pan only and it tips. Take 4 off both pans and it stays level, with $x$ alone on the left and 7 on the right.

The four one-step equations use four pairs of inverse operations:

• $x + p = q$: subtract $p$ from both sides.
• $x - p = q$: add $p$ to both sides.
• $px = q$: divide both sides by $p$.
• $x \div p = q$: multiply both sides by $p$.

5. What a solution is, and how to test one

An equation is a question with a yes-or-no test. To test a number, substitute it for the variable and work out both sides. If they are equal, the number is a solution; if not, it is not.

Which of 3, 4 and 5 solves $6y = 24$? Substitute each: $6 \times 3 = 18$, no; $6 \times 4 = 24$, yes; $6 \times 5 = 30$, no. So $y = 4$.

Testing is also how you check an answer you found by undoing. Always put your solution back into the original equation, not into a later line, because a mistake in an early line would be copied into the later ones.

The equations in this lesson each have exactly one solution. That is different from an inequality such as $y > 4$, which is true for many numbers. An equation pins the variable down to one value; an inequality allows a whole stretch of the number line.

6. Adding and subtracting: x + p = q and x − p = q

When a number is added to the variable, subtract it from both sides. When a number is subtracted from the variable, add it to both sides.

For $n + 2.6 = 9.1$: subtract 2.6 from both sides, so $n = 9.1 - 2.6 = 6.5$. Check: $6.5 + 2.6 = 9.1$.

For $k - 18 = 45$: add 18 to both sides, so $k = 45 + 18 = 63$. Check: $63 - 18 = 45$.

A size check catches most mistakes. In $x + p = q$ with positive numbers, $x$ must be smaller than $q$, because something was added to reach $q$. In $x - p = q$, $x$ must be larger than $q$, because something was taken away to reach $q$. If your answer breaks this, you used the wrong operation.

7. Multiplying and dividing: px = q and x ÷ p = q

When the variable is multiplied by a number, divide both sides by it. $8m = 60$ gives $m = 60 \div 8 = 7.5$. The answer does not have to be a whole number; decimals and fractions are fine.

When the variable is divided by a number, multiply both sides by it. $t \div 5 = 3.2$ gives $t = 3.2 \times 5 = 16$.

When the coefficient is a fraction, multiply both sides by its reciprocal. For $\tfrac{2}{5}w = 14$, multiply by $\tfrac{5}{2}$: $w = 14 \times \tfrac{5}{2} = \tfrac{70}{2} = 35$. This works because $\tfrac{5}{2} \times \tfrac{2}{5} = 1$, so the left side becomes $1w$, which is just $w$. Dividing by $\tfrac{2}{5}$ and multiplying by $\tfrac{5}{2}$ are the same move, as you learned when dividing fractions.

A coefficient written as a decimal works the same way: $0.25a = 3$ gives $a = 3 \div 0.25 = 12$.

8. When the variable is on the right

Equations do not always put the variable on the left. $20 = x + 8$ says the same thing as $x + 8 = 20$, because 'equals' works in both directions: if one amount equals another, the second equals the first. Solve it exactly the same way. Subtract 8 from both sides to get $12 = x$, which is the same as $x = 12$.

Many people like to flip such an equation around first so the variable is on the left, and that is always allowed. What is not allowed is to move only part of a side. In $20 = x + 8$ the whole right side, $x + 8$, is one amount; the 8 cannot be moved across without subtracting it from both sides. Keeping the two sides whole, and changing them together, is the one idea behind every equation you will solve, in this grade and in every grade after it.

9. Writing an equation from a story

Real problems arrive in words, not symbols. To turn words into an equation:

1. Choose a letter for the unknown and say what it stands for, with its unit: 'let $h$ be the number of hours'.
2. Find the sentence that says two amounts are equal: 'costs', 'is', 'totals', 'leaves'.
3. Write each amount as an expression and put an equals sign between them.

'A movie ticket costs 12.50 dollars, and a group paid 87.50 dollars' becomes $12.5t = 87.5$, where $t$ is the number of tickets. 'After a 15-foot drop, the kite was 42 feet high' becomes $k - 15 = 42$, where $k$ is the starting height. Once the equation is written, solving it is the easy part. Finish by answering the question in a sentence with the unit: '7 tickets were bought'.

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

To solve a one-step equation:

1. Find the operation that was done to the variable: added, subtracted, multiplied or divided, and by what number.
2. Choose the inverse: subtract for added, add for subtracted, divide for multiplied, multiply for divided. For a fraction coefficient, multiply by the reciprocal.
3. Do it to both sides, and write the new equation.
4. Simplify each side until the variable stands alone.
5. Answer the question, with a unit if there is a story.

To check:

• Substitute your solution into the original equation. Both sides must come out equal.
• Use the size check: added-to means smaller than the right side; subtracted-from means larger; multiplied by more than 1 means smaller; divided means larger.
• Read the answer in the story. A number of tickets must be a whole number; a price must be a sensible amount of money.

11. In the world: the federal tax on gasoline

Every gallon of gasoline sold in the United States carries a federal tax of 18.4 cents, which is 0.184 dollars; states add their own taxes on top. The money pays for highways and bridges. Suppose a delivery driver's receipt shows that the federal tax on one fill-up was 2.76 dollars. How many gallons did the driver buy? Let $g$ be the number of gallons. Then

$$0.184g = 2.76.$$

Divide both sides by 0.184: $g = 2.76 \div 0.184 = 15$ gallons. Check: $0.184 \times 15 = 2.76$. The same equation works in reverse for a trucking company: a truck that burns 20,000 gallons in a year pays $0.184 \times 20{,}000 = 3{,}680$ dollars of federal gasoline tax. One-step equations let an accountant move between the tax paid and the fuel bought.

12. In the world: how far is left in a marathon

A marathon is 26.2 miles long. A runner's watch shows that she has 9.7 miles to go. How far has she already run? Let $r$ be the miles run so far. The miles run plus the miles left make the whole race:

$$r + 9.7 = 26.2.$$

Subtract 9.7 from both sides: $r = 26.2 - 9.7 = 16.5$ miles. Check: $16.5 + 9.7 = 26.2$. If she has been running at a steady 10 minutes per mile, the time so far is another one-step equation, $t \div 10 = 16.5$, where $t$ is minutes, so $t = 165$ minutes, which is 2 hours and 45 minutes. Runners and race organizers set up equations like these all the time to plan water stations and to predict finishing times.

13. Mistakes to watch for

Doing the same operation instead of the inverse. For $x + 5 = 12$, adding 5 gives $x + 10 = 17$, which does not help. Subtract 5.

Changing only one side. Subtracting 5 from the left alone breaks the balance: $x = 12$ is wrong.

Dividing the wrong way round. For $4y = 10$, $y = 10 \div 4 = 2.5$, not $4 \div 10$.

Multiplying by the fraction instead of its reciprocal. For $\tfrac{2}{3}x = 12$, $x = 12 \times \tfrac{3}{2} = 18$, not $12 \times \tfrac{2}{3} = 8$.

Skipping the check. Substituting takes ten seconds and catches every one of these mistakes.

14. Undoing an addition with decimals

1. Solve $x + 7.5 = 12$. Name the operation done to $x$.

   $7.5 \text{ was added to } x$

   To get x alone, undo what was done to it.

2. Choose the inverse operation.

   $\text{subtract } 7.5$

   Subtracting 7.5 undoes adding 7.5.

3. Subtract 7.5 from both sides.

   $x + 7.5 - 7.5 = 12 - 7.5$

   Doing the same to both sides keeps the equation balanced.

4. Simplify both sides.

   $x = 4.5$

   On the left the 7.5s cancel; on the right, 12 minus 7.5 is 4.5.

5. Check in the original equation.

   $4.5 + 7.5 = 12$

   Both sides are equal, so 4.5 is the solution.

15. A fraction coefficient

1. Solve $\tfrac{3}{4}w = 18$. Name the operation done to $w$.

   $w \text{ was multiplied by } \tfrac{3}{4}$

   The fraction in front of the variable is its coefficient.

2. Choose the inverse: multiply by the reciprocal.

   $\tfrac{3}{4} \to \tfrac{4}{3}$

   A number times its reciprocal is 1, which leaves w alone.

3. Multiply both sides by $\tfrac{4}{3}$.

   $\tfrac{4}{3} \times \tfrac{3}{4}w = \tfrac{4}{3} \times 18$

   Both sides change in the same way, so the balance holds.

4. Simplify the right side.

   $\tfrac{4 \times 18}{3} = \tfrac{72}{3} = 24, \quad w = 24$

   Multiply the top, then divide by the bottom.

5. Check in the original equation.

   $\tfrac{3}{4} \times 24 = 18$

   A quarter of 24 is 6, and three quarters is 18.

6. Check the size of the answer.

   $24 > 18$

   Multiplying by a fraction less than 1 made w smaller, so w must be bigger than 18.

16. Writing and solving an equation from a story

1. Liam spent 13.75 dollars on a book and has 26.40 dollars left. How much did he have before? Choose a variable.

   $s = \text{dollars Liam had before}$

   Naming the unknown, with its unit, is the first step of any story problem.

2. Write the equation from the story.

   $s - 13.75 = 26.40$

   What he had, minus what he spent, is what is left.

3. Name the operation done to $s$.

   $13.75 \text{ was subtracted from } s$

   Spending takes money away.

4. Add 13.75 to both sides.

   $s - 13.75 + 13.75 = 26.40 + 13.75$

   Adding undoes subtracting.

5. Add the decimals.

   $26.40 + 13.75 = 40.15, \quad s = 40.15$

   Line up the decimal points: hundredths with hundredths.

6. Check in the original equation.

   $40.15 - 13.75 = 26.40$

   Spending 13.75 dollars from 40.15 dollars leaves 26.40 dollars.

7. Answer in a sentence.

   $\text{Liam had } 40.15 \text{ dollars}$

   He had more before spending than after, as he should.

17. Your turn: solve $x \div 6 = 4.5$

1. Name the operation done to x.

   $x \text{ was divided by } 6$

   The inverse of dividing is multiplying.

2. Multiply both sides by 6.

   $x \div 6 \times 6 = 4.5 \times 6$

   Do the same to both sides.

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

   Simplify the right side.

18. Guided practice

Match each equation with the number that makes it true.

19. Guided practice

Solve $\dfrac{5}{8}x = 10$. Complete the worked solution.

1. Name the operation done to x.

   $x \text{ was multiplied by } \dfrac{5}{8}$

   The fraction written in front of x is a factor.

2. Multiply the right side by 8, the bottom of the fraction.

   $10 \times 8 =$ _t

   Multiplying both sides by the reciprocal $\dfrac{8}{5}$ starts with multiplying by 8.

3. Divide that product by 5, the top of the fraction.

   $\text{product} \div 5 =$ _x

   Dividing by 5 finishes the multiplication by the reciprocal; on the left only x is left.

4. Check by substituting into the original equation.

   $\dfrac{5}{8} \times x = 10$

   Taking $\dfrac{5}{8}$ of the solution must give back the right side.

20. Guided practice

Solve $x + 10 = 24$. Write one equation per line, ending with $x = $ your answer.

x + 10 = 24

21. Practice

Solve $y - 2.9 = 8.8$. Which number do you add to both sides, and what is y?

Add _add to both sides, so y = _answer.

22. Practice

Solve $5y = 33$. Give the division you do on the right side, and y as a decimal.

Divide 33 by _div, so y = _answer.

23. Practice

Solve both equations: $m \div 7 = 5$ and $n \div 7 = 9$.

m = _m and n = _n

24. Somewhere new

Sam bought some stickers at 9 dollars each and paid 45 dollars in all. Call the number of stickers p. Write the equation and solve it.

The equation is 9p = _total, so Sam bought _answer stickers.

25. Lesson test

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

26. Test question

A 6-pack of lemonade costs 11.52 dollars. Write an equation for p, the price of one bottle in dollars, and solve it. At that price, what would 11 single bottles cost?

One bottle costs _p dollars, and 11 bottles cost _cost dollars.

27. What you can do now

You can solve and check one-step equations. Without looking: how do you solve $\tfrac{2}{3}x = 12$, and how do you check the answer?

Working for the steps left to you

17. Your turn: solve $x \div 6 = 4.5$, step 3