Presentation is loading. Please wait.

Presentation is loading. Please wait.

Math 025 Unit 5 Section 6.7.

Similar presentations


Presentation on theme: "Math 025 Unit 5 Section 6.7."— Presentation transcript:

1 Math Unit 5 Section 6.7

2 A literal equation is an equation that contains more than one variable.
Examples: 2x + 3y = 6 4w – 2x + z = 7 Many formulas are literal equations. A = l w V = lwh To solve a literal equation, use equation solving techniques to rearrange the equation until the desired variable is alone on one side of the equation.

3 Objective: To solve a literal equation
Solve for y: 3x – 4y = 12 Subtract 3x from both sides -4y = 12 – 3x Divide both sides by -4 y = 12 – 3x The answer can be rearranged if desired y = 3x –

4 Objective: To solve a literal equation
Solve for x: 5x – 2y = 10 Add 2y to both sides 5x = y Divide both sides by 5 x = y The answer can be rearranged if desired 2y x = + 2 5

5 Objective: To solve a literal equation
Solve for R: I = Multiply the equation by (R + r) to clear any fractions E (R + r) R + r IR + Ir = E Subtract Ir from both sides IR = E – Ir Divide both sides by I R = E – Ir I

6 Objective: To solve a literal equation
Solve for c: L = a(1 + c) Simplify the right side by multiplying L = a + ac Subtract a from both sides L – a = ac Divide both sides by a c = L – a a

7 Objective: To solve a literal equation
Solve for C: S = C – rC Factor the right side S = C(1 – r) Divide both sides by (1 – r) C = S – r


Download ppt "Math 025 Unit 5 Section 6.7."

Similar presentations


Ads by Google