## Key Concepts

- Create linear equations and use them to solve problems
- Solve consecutive integer problems
- Use equations to solve problems
- Solve work and time problems.

### Introduction to Solving Linear Equation

Understand Solving Linear Equations and solve the given problem:

- Sam saves $12 dollars each month. How many will it take Sam to save a total of $144?

**Solution: **

Let the number of months that Sam saved money be x.

Total money saved by Sam in x months = x × 12

Given: Money saved = $144

⇒ 144=12x

⇒ x = 144/12

⇒ x = 12

Therefore, Sam saved money for 12 months.

**Example of solving a linear equation to find an unknown value:**

Solve for “a” if 3(a+2) − 7(1−a) = 13

Sol: 3(a+2) − 7(1−a) = 13

3a+6 − 7+7a = 13

10a−1 = 13

10a = 14

a = 14 / 10

a = 7 / 5

### When do we use linear equations?

- To solve problems that include consecutive integers.
- To solve work and time related problems.
- To find the unknown value.
- To solve mixture problems.

### Solving consecutive integer problems using linear equations

**We can use a linear equation to find an unknown number when the sum or product or difference between consecutive numbers is given.**

**Example:** The sum of three consecutive integers is 96. Find the three integers.

Sol: Let the smallest number be p.

The next two consecutive numbers will be (p+1) and (p+2).

Given: p+(p+1) +(p+2) = 96

3p + 3 = 96

3p = 93

p = 93 / 3

p = 31

The next number is (p+1) = 31+1

= 32

The next number is (p+2) = 31+2

= 33

Therefore, the three consecutive numbers are 31, 32, and 33.

### Use linear equations to solve mixture problems

**Example:** If the lab technician needs 30 liters of a 25% acid solution, how many liters of the 10% and the 30% acid solutions should she mix to get what she needs?

Sometimes, one value depends on the other. Then, we assume a variable for the first value and substitute in the second value. Write an equation relating the number of litres of acid in each solution.

Let the total number of liters of one solution be x

The number of liters of other solution will be 30 – x

Now, 25% solution of 30L = 10% solution of x + 30% solution of (30 – x)

25 / 100×30 = 10 / 100×x + 30 / 100×(30−x)

0.25×30 = 0.1×x+0.3×(30−x)

### Solve work and time problems

Sometimes, we need to use linear equations to solve work and time related problems.

**Example:** A kid swims at a rate of 0.9 miles per hour. Then he takes a break for 30 minutes and swims again at a rate of 1.2 miles per hour. The length of the swimming pool is 36 miles.

**Sol:** Let the total time taken by the kid be “t” hours.

Time taken to swim from start to end + Break time + Time taken to swim from end to start = t

36 miles / 0.9 miles per hour+30 minutes + 36 miles / 1.2 miles per hour = t

40+30+30 = t

t = 100 minutes

So, the kid completes the ride in 100 minutes.

## Exercise

Solve each equation.

- -4x+3x = 2
- 7 = 5y-13-y
- 7m-4-9m-36 = 0
- -2 = -5t + 10+2t

Solve each equation.

- 2(2x+1) = 26
- -2(2z+ 1) = 26
- 92 = -4(2r-5)

### Concept map

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