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Linear Equations Sections 10.1 – 10.3.

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Presentation on theme: "Linear Equations Sections 10.1 – 10.3."— Presentation transcript:

1 Linear Equations Sections 10.1 – 10.3

2 I can recognize linear equations.
I can solve equations of the form x + B = C. I can solve equations of the form Ax = C I can solve equations of the form Ax + B = C

3 Write an Equation for the Following
I added $30 to my bank account and my balance is $330. What did I start with? I paid $50 for 10 bags of cherries. How much did each bag cost? Parking costs a flat rate of $3.00 plus $2.00 per hour. I spent $ How long was I parked?

4 These Are Linear Equations
x + 30 = 330 10x = 50 2x + 3 = 13 Any equation that can be written in the form Ax + B = C, where A, B, C are real numbers.

5 NOT Linear equations x2 + 5x -3 = 0 |x - 3| = 7 1/x = 12 √x = 25.

6 Linear or Nonlinear? 5 = 2x 3 – s = ¼ 3 – t2 = ¼ 50 = ¼ r2

7 Goal: Solve Linear Equations
We have simplified expressions with one or sometimes more than one variable. Today we are going to learn how to solve linear equations in one variable.

8 This means: what value of x makes the sentence true?
Try This Problem I deposited $30 into my bank account and my new balance was $330. What did I start with? How did you figure this out? Try to solve this equation: x + 3 = 7. This means: what value of x makes the sentence true?

9 Simplifying Expressions
When you simplify expressions, you only have one side of a scale. The weight cannot change. 2(x+3) 2x+6

10 Solving Equations When you solve equations, you have both sides of the scale! You can change the weight, but the scale must balance! x + 3 = 7 x = 7 - 3 x = 4

11 Inverse Operations To solve x + 3 = 7, you subtract 3 from both sides.
To solve x - 3 = 7, what do you do? Addition and subtraction are inverse operations. We always use inverse operations to solve equations.

12 Always check your answer!
More Examples If -x = a, then x = -a 5 – k = 12 5m + 4 = 6m y + 2/3 = ½ ½ x – 5 = -1/2 x + 2 6m – 5m = m (like terms) Always check your answer!

13 Don’t forget to check your answer!
Try Some x – 17 = 25 12 – r = 7 t – ½ = 3/4 Don’t forget to check your answer!

14 Use the distributive property!
Simplify First Solve 5t – 4t + 6 = 9 Simplify first, then solve Solve 4x x – 3 = 9 + 5x – 4 Solve: 3(2+5x) – (1+14x) = 6 5t – 4t = t t + 6 = 9 Use the distributive property!

15 What About Multiplication?
If I paid $50 for 10 bags of cherries, how much did each bag of cherries cost? How did you figure this out? Solve: 5x = 60

16 Try Some More Solve -25p = 50 Solve 2m = 15 Solve -6x = 14.

17 Fractions I In algebra, instead of writing x ¥ 3, we write x/3.
This is consistent with the notion that division is multiplication by the reciprocal. To solve x/3 = 10, how do we undo the division?

18 Fractions II What about the equation 2/3 x = 6?
To divide by 2/3, multiply by the reciprocal. Since (3/2) £ (2/3) = 1, we now have: x = (3/2) £ 6. Work this out, what do we get? x = 9

19 Let’s do Another Together
Solve: 7/5 x = 9/4 What is the reciprocal of 7/5? Multiply both sides by 5/7. What do we have on the left? For the right, what is 5/7 times 9/4? What is our solution? 5/7 Just x The product of a number with its reciprocal is 1! 45/28 x = 45/28

20 Try These Solve y/12 = 5 Solve 1/3 z = 19 Solve 6/7 t = 9/5

21 Simplify First Just like before, sometimes you have to simplify before solving! Solve: 5x + 6x = 9 Solve 7x – 2x = -25

22 Let’s Put These Together!
Parking costs a flat rate of $3.00 plus $2.00 per hour. I spent $ How long was I parked? Answer the question with your partner. How did you solve this problem?

23 The Easiest Way To Do This
The equation for this problem is: 2h + 3 = 13. The variable h stands for hours. First: subtract the 3 from both sides: 2h = 10 Second: divide both sides by 2: h = 5.

24 The order is the opposite!
Another example The order is the opposite! Solve 3x -5 = 7 First: addition and subtraction Second: multiplication and division Add 5 to both sides 3x = 12 Divide both sides by 3 x = 4

25 Try These: Solve: 5x – 6 = 17 Solve: -4x + 2 = 9

26 You can solve linear equations using inverse operations.
If you have to deal with both addition and multiplication, deal with addition first.


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