Equations – Success if you can do these 3x = 64x + 2 = 3x + 7 x/4 = 25x – 3 = 7 + 3x x + 3 = 173(x + 2) = 12 x – 4 = 134(x + 5) = 32 3x + 4 = 253(2x –

Slides:

Advertisements

Similar presentations

Equations with the unknown on both sides.

Advertisements

Equations – Success if you can do these 3x = 64x + 2 = 3x + 7 x/4 = 25x – 3 = 7 + 3x x + 3 = 173(x + 2) = 12 x – 4 = 134(x + 5) = 32 3x + 4 = 253(2x –

Solving Equations = 4x – 5(6x – 10) -132 = 4x – 30x = -26x = -26x 7 = x.

Solve an equation with variables on both sides

Solve an absolute value equation EXAMPLE 2 SOLUTION Rewrite the absolute value equation as two equations. Then solve each equation separately. x – 3 =

Solve an equation using subtraction EXAMPLE 1 Solve x + 7 = 4. x + 7 = 4x + 7 = 4 Write original equation. x + 7 – 7 = 4 – 7 Use subtraction property of.

Equations & Brackets.. You are now going to solve more complex equations by combining together two ideas that you have seen already. Try the following.

Solving Percent Problems Using Proportions

Standardized Test Practice

Solving Equations with Brackets or Fractions. Single Bracket Solve 3(x + 4) = 24 3x + 12 = 24 3x + 12 – 12 = x = 12 x = 4 Multiply brackets out.

Solve a radical equation

Solve the equation -3v = -21 Multiply or Divide? 1.

 Sometimes you have a formula and you need to solve for some variable other than the "standard" one. Example: Perimeter of a square P=4s It may be that.

Solving Equations Using Multiplication and Division Algebra 1 Section 3.2a.

Section 3.4: Solving Equations with Variables on Both Sides Algebra 1 Ms. Mayer.

Systems of Equations 7-4 Learn to solve systems of equations.

Factoring Polynomials by Completing the Square. Perfect Square Trinomials l Examples l x 2 + 6x + 9 l x x + 25 l x x + 36.

Solving Systems of Equations: The Elimination Method Solving Systems of Equations: The Elimination Method Solving Systems of Equations: The Elimination.

Solve an equation using addition EXAMPLE 2 Solve x – 12 = 3. Horizontal format Vertical format x– 12 = 3 Write original equation. x – 12 = 3 Add 12 to.

Example 1 Solving Two-Step Equations SOLUTION a. 12x2x + 5 = Write original equation. 112x2x + – = 15 – Subtract 1 from each side. (Subtraction property.

Solving Linear Equations Substitution. Find the common solution for the system y = 3x + 1 y = x + 5 There are 4 steps to this process Step 1:Substitute.

Use the substitution method

Solve Equations With Variables on Both Sides. Steps to Solve Equations with Variables on Both Sides  1) Do distributive property  2) Combine like terms.

Simple Equations Brackets Consider the simple equation 5(2x+3) = 3(x+9) Remove the brackets to obtain 10x + 15 = 3x + 27 Subtract 15 from both sides.

2.4 – Solving Equations with the Variable on Each Side.

* Collect the like terms 1. 2a = 2a x -2x + 9 = 6x z – – 5z = 2z - 6.

Solving 2 step equations. Two step equations have addition or subtraction and multiply or divide 3x + 1 = 10 3x + 1 = 10 4y + 2 = 10 4y + 2 = 10 2b +

Solving Equations with Variables on Both Sides. Review O Suppose you want to solve -4m m = -3 What would you do as your first step? Explain.

Lesson 7.4 Solving Multiplication and Division Equations 2/3/10.

4-6 Solving Equations with Fractions What You’ll Learn ► To solve one – step equations ► To solve two – equations.

Substitution Method: Solve the linear system. Y = 3x + 2 Equation 1 x + 2y=11 Equation 2.

Jeopardy Solving Equations

Solving Multi-Step Equations

Equations Pre/Post Test

Solving Two-Step Equations

Solve a quadratic equation

Solving Two and Multi Step Equations

Example 2 4 m 8 m 5m 12 m x y.

Solving Two Step Equations

Solving Inequalities.

Solving Multi-Step Equations

Example 2 4 m 8 m 5m 12 m x y.

Solving Multi-Step Equations

Do Now 1) t + 3 = – 2 2) 18 – 4v = 42.

Solving Two-Step Equations

Equations – Success if you can do these

Solving Multi-Step Equations

Solving Multi-Step Equations

Solving one- and two-step equations

a + 2 = 6 What does this represent? 2 a

Solving Equations by Combining Like Terms

Equations – Success if you can do these

Solving Multi-Step Equations

Solving Equations with Variables on Both Sides

Solving Equations with Variables on Both Sides

Think about… How is an equation like a balance scale?

Solving Multi-Step Equations

Equations – Success if you can do these

Math-7 NOTES What are Two-Step Equations??

Solving Two Step Algebraic Equations

5a + 3 = 2a a = 2a a = 2a + 9 5a – 2a = 9 3a = 9 a = 9

Do Now Solve. 1. –8p – 8 = d – 5 = x + 24 = 60 4.

Do Now Solve. n + 9 = z = 55 n = x = z = -16

Equations – Success if you can do these

Solving basic equations

Solving Linear Equations

Solving Algebraic Equations with Addition and Subtraction

Objective Solve radical equations.. Objective Solve radical equations.

Equations – Success if you can do these

Solving Algebraic Equations with Addition and Subtraction

Presentation transcript:

Equations – Success if you can do these 3x = 64x + 2 = 3x + 7 x/4 = 25x – 3 = 7 + 3x x + 3 = 173(x + 2) = 12 x – 4 = 134(x + 5) = 32 3x + 4 = 253(2x – 2) = 2(x + 6) 4x – 3 = 174(3x + 5) = 2(x + 15)

How to solve the 4(x + 3) = 40 type of equation The 4, 3 and 40 could be any numbers

Get rid of the brackets 4(x + 3) = 40

Get rid of the brackets 4(x + 3) = 40 X x x x + 12 = 40

Get rid of the brackets 4(x + 3) = 40 X x x x + 12 = 40 Subtract 12 from both sides

Get rid of the brackets 4(x + 3) = 40 X x x x + 12 = 40 4x + 12 – 12 = 40 – 12 Subtract 12 from both sides

Get rid of the brackets 4(x + 3) = 40 X x x x + 12 = 40 4x + 12 – 12 = 40 – 12 4x =28 Subtract 12 from both sides

Get rid of the brackets 4(x + 3) = 40 X x x x + 12 = 40 4x + 12 – 12 = 40 – 12 4x =28 Subtract 12 from both sides Divide both sides by 4

Get rid of the brackets 4(x + 3) = 40 X x x x + 12 = 40 4x + 12 – 12 = 40 – 12 4x =28 4x/4 = 28/4 Subtract 12 from both sides Divide both sides by 4

Get rid of the brackets 4(x + 3) = 40 X x x x + 12 = 40 4x + 12 – 12 = 40 – 12 4x =28 4x/4 = 28/4 = 7 Subtract 12 from both sides Divide both sides by 4

Solve this equation 5(x + 2) = 3(x + 10)

Get rid of the brackets

5(x + 2) = 3(x + 10) 5x + 10 = 3x + 30 Get rid of the brackets

5(x + 2) = 3(x + 10) 5x + 10 = 3x x + 10 –10 = 3x + 30 –10 Get rid of the brackets Subtract 10 from both sides

5(x + 2) = 3(x + 10) 5x + 10 = 3x x + 10 –10 = 3x + 30 –10 5x = 3x + 20 Get rid of the brackets Subtract 10 from both sides

5(x + 2) = 3(x + 10) 5x + 10 = 3x x + 10 –10 = 3x + 30 –10 5x = 3x + 20 Get rid of the brackets Subtract 10 from both sides Subtract 3x from both sides

5(x + 2) = 3(x + 10) 5x + 10 = 3x x + 10 –10 = 3x + 30 –10 5x = 3x x – 3x = 3x + 20 –3x 2x = 20 Get rid of the brackets Subtract 10 from both sides Subtract 3x from both sides

5(x + 2) = 3(x + 10) 5x + 10 = 3x x + 10 –10 = 3x + 30 –10 5x = 3x x – 3x = 3x + 20 –3x 2x = 20 Get rid of the brackets Subtract 10 from both sides Subtract 3x from both sides Divide both sides by 2

5(x + 2) = 3(x + 10) 5x + 10 = 3x x + 10 –10 = 3x + 30 –10 5x = 3x x – 3x = 3x + 20 –3x 2x = 20 2x/2 = 20/2 x = 10 Get rid of the brackets Subtract 10 from both sides Subtract 3x from both sides Divide both sides by 2

Remember, always do the same thing to both sides. Felix is checking that you do this properly. 5x + 3 = x 5x + 3 – 3 = 43 – 3 5x = x 5x – 3x = x – 3x 2x = 40 x = 20