Find lengths using Theorem 10.14

Slides:

Advertisements

Similar presentations

( ) EXAMPLE 3 Solve ax2 + bx + c = 0 when a = 1

Advertisements

EXAMPLE 2 Use a ratio to find a dimension SOLUTION Painting You are planning to paint a mural on a rectangular wall. You know that the perimeter of the.

Solving Multiplication and Division Equations. EXAMPLE 1 Solving a Multiplication Equation Solve the equation 3x = x = 15 Check 3x = 45 3 (15)

EXAMPLE 4 Solve proportions SOLUTION a x 16 = Multiply. Divide each side by 10. a x 16 = = 10 x5 16 = 10 x80 = x8 Write original proportion.

Solve an equation with variables on both sides

Find hypotenuse length in a triangle EXAMPLE 1

Solve an equation by combining like terms

EXAMPLE 1 Solve a quadratic equation having two solutions Solve x 2 – 2x = 3 by graphing. STEP 1 Write the equation in standard form. Write original equation.

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

EXAMPLE 2 Using the Cross Products Property = 40.8 m Write original proportion. Cross products property Multiply. 6.8m 6.8 = Divide.

EXAMPLE 1 Find hypotenuse length in a triangle o o o Find the length of the hypotenuse. a. SOLUTION hypotenuse = leg 2 = 8 2 Substitute

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.

In the skateboard design, VW bisects XY at point T, and XT = 39.9 cm. Find XY. Skateboard SOLUTION EXAMPLE 1 Find segment lengths Point T is the midpoint.

Standardized Test Practice

Step 1: Simplify Both Sides, if possible Distribute Combine like terms Step 2: Move the variable to one side Add or Subtract Like Term Step 3: Solve for.

EXAMPLE 1 Collecting Like Terms x + 2 = 3x x + 2 –x = 3x – x 2 = 2x 1 = x Original equation Subtract x from each side. Divide both sides by x2x.

Standardized Test Practice

EXAMPLE 1 Find the length of a hypotenuse SOLUTION Find the length of the hypotenuse of the right triangle. (hypotenuse) 2 = (leg) 2 + (leg) 2 Pythagorean.

EXAMPLE 3 Solve an equation by factoring Solve 2x 2 + 8x = 0. 2x 2 + 8x = 0 2x(x + 4) = 0 2x = 0 x = 0 or x + 4 = 0 or x = – 4 ANSWER The solutions of.

Standardized Test Practice

Algebra I Chapter 10 Review

( ) EXAMPLE 3 Standardized Test Practice SOLUTION 5 x = – 9 – 9

Solving Two-Step Equations You will solve equations by undoing operations using properties of equality. Essential Question: How do you solve two-step equations?

EXAMPLE 2 Rationalize denominators of fractions Simplify

Solve a logarithmic equation

EXAMPLE 4 Solve a logarithmic equation Solve log (4x – 7) = log (x + 5). 5 5 log (4x – 7) = log (x + 5) x – 7 = x x – 7 = 5 3x = 12 x = 4 Write.

EXAMPLE 2 Rewrite a formula with three variables SOLUTION Solve the formula for w. STEP 1 P = 2l + 2w P – 2l = 2w P – 2l 2 = w Write perimeter formula.

EXAMPLE 3 Find lengths using Theorem Use the figure at the right to find RS. SOLUTION 256= x 2 + 8x 0= x 2 + 8x – 256 RQ 2 = RS RT 16 2 = x (x +

Ratio and Proportion 7-1.

Solve an equation by combining like terms EXAMPLE 1 8x – 3x – 10 = 20 Write original equation. 5x – 10 = 20 Combine like terms. 5x – =

EXAMPLE 3 Find possible side lengths ALGEBRA

EXAMPLE 3 Find possible side lengths ALGEBRA A triangle has one side of length 12 and another of length 8. Describe the possible lengths of the third side.

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

Example 3 Solving an Equation Using Addition The solution is ANSWER Original equation 13=4.5c– Add 4.5 to each side. (Addition property of equality)

Standardized Test Practice EXAMPLE 1. SOLUTION Standardized Test Practice Write and solve a two-step equation to find the number of flamingos. Write a.

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 Solve by equating exponents Rewrite 4 and as powers with base Solve 4 = x 1 2 x – 3 (2 ) = (2 ) 2 x – 3x – 1– 1 2 = 2 2 x– x + 3 2x =

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.

EXAMPLE 2 Checking Solutions Tell whether (7, 6) is a solution of x + 3y = 14. – x + 3y = 14 Write original equation ( 6) = 14 – ? Substitute 7 for.

EXAMPLE 1 Solve a two-step equation Solve + 5 = 11. x 2 Write original equation. + 5 = x – 5 = x 2 11 – 5 Subtract 5 from each side. = x 2 6 Simplify.

Use the substitution method

Algebra Core Review Day 8. Unit 17: Radicals How do you simplify radicals? When you add or subtract radicals they must be _______! How do you multiply.

Solve a two-step equation by combining like terms EXAMPLE 2 Solve 7x – 4x = 21 7x – 4x = 21 Write original equation. 3x = 21 Combine like terms. Divide.

EXAMPLE 1 Use the Perpendicular Bisector Theorem AD = CD Perpendicular Bisector Theorem 3x x =5x = Substitute. 7 x = Solve for x. BD is the perpendicular.

Warm-Up Exercises ANSWER –1, –3 1.x 2 + 4x + 3 = 0 ANSWER 3 2.6x – 9 = x 2 Solve the equation.

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

Find Segment Lengths in Circles

1. Add: 5 x2 – 1 + 2x x2 + 5x – 6 ANSWERS 2x2 +7x + 30

Rewrite a formula with three variables

EXAMPLE 2 Rationalize denominators of fractions Simplify

Solve a literal equation

( ) EXAMPLE 3 Standardized Test Practice SOLUTION 5 x = – 9 – 9

Solve an equation by multiplying by a reciprocal

Solve the equation. 1. x2 + 4x + 3 = x – 9 = x2 ANSWER –1, –3

6-2 Solving Systems Using Substitution

Solving Systems Using Substitution

Use properties of parallelograms

EXAMPLE 4 Apply Theorem 8.19 Find m D in the kite shown at the right.

Solve an equation by combining like terms

Solving Multi-Step Equations

Find an unknown interior angle measure

- Finish Unit 1 test - Solving Equations variables on both sides

Standardized Test Practice

Find lengths using Theorem 10.14

Solve an inequality using subtraction

Segment Lengths in Circles

3-2 Solving Inequalities Using Addition and Subtraction

EXAMPLE 4 Solve proportions Solve the proportion. ALGEBRA a x 16

Presentation transcript:

Find lengths using Theorem 10.14 EXAMPLE 1 Find lengths using Theorem 10.14 ALGEBRA Find ML and JK. SOLUTION NK NJ = NL NM Use Theorem 10.14. x (x + 4) = (x + 1) (x + 2) Substitute. x2 + 4x = x2 + 3x + 2 Simplify. 4x = 3x + 2 Subtract x2 from each side. x = 2 Solve for x.

EXAMPLE 1 Find lengths using Theorem 10.14 Find ML and JK by substitution. SOLUTION ML = ( x + 2 ) + ( x + 1) JK = x + ( x + 4) = 2 + 2 + 2 + 1 = 2 + 2 + 4 = 7 = 8

Standardized Test Practice EXAMPLE 2 Standardized Test Practice SOLUTION RQ RP = RS RT Use Theorem 10.15. 4 (5 + 4) = 3 (x + 3) Substitute. 36 = 3x + 9 Simplify. 9 = x Solve for x ANSWER The correct answer is D.

GUIDED PRACTICE for Examples 1 and 2 Find the value(s) of x. SOLUTION = 6 (6 + 9) Use Theorem 10.15. 90 = 5x + 25 Simplify. 13 = x Solve for x

GUIDED PRACTICE for Examples 1 and 2 Find the value(s) of x. SOLUTION = (4) (6) Substitute. 3x = 24 Simplify. x = 8 Solve for x.

GUIDED PRACTICE for Examples 1 and 2 Find the value(s) of x. SOLUTION = (x+1)(x + 1 + x – 1) Use Theorem 10.15. 3x + 15 = 2x2 + 2x Simplify. 0 = 2x2 – x – 15 Combine Like Terms. 0 = (x – 3) (2x + 5) Solve for x 3 = x Use positive solution