Literal Equations and Formulas Honors Math – Grade 8.

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Presentation transcript:

Literal Equations and Formulas Honors Math – Grade 8

Get Ready for the Lesson Suppose the designer of the Magnum XL-200 decided to adjust the height of the second hill so that the coaster would have a speed of 49 feet per second when it reached the top. If we ignore friction, the equation below can be used to find the height of the second hill. In this equation, g represents the acceleration due to gravity, h is the height of the second hill, and v is the velocity of the coaster when it reaches the top of the second hill. Some equations such as this one contain more than one variable. It is often useful to solve these equations for one of the variables. In this lesson, we will explore how to solve for a specific variable.

Solve each equation for the specified variable. Since the problem states to solve for y, do not move that variable. Move 3x to the other side. Subtract 3x from both sides of the equation. Divide both sides of the equation by -4. Since the denominator has a negative integer, you may distribute a -1 to the numerator to remove it from the denominator.

Solve each equation for the specified variable. Since the problem states to solve for n, do not move that variable. Move 6p to the other side. Subtract 6p from both sides of the equation. Divide both sides of the equation by -4. Since the coefficients in the numerator are divisible by 3, divide them by 3 to simplify the expression further.

Solve each equation for the specified variable. Since the problem states to solve for k, do not move that variable. Since the equation contains a grouping symbol, use inverse operations to eliminate the denominator. Multiply both sides of the equation by 5. Add 2 to both sides of the equation.

Solve each equation for the specified variable. Since the problem states to solve for m, move the terms that contain the variable to the same side. Move the other terms to the other side. Move sm to the other side of the equation. Subtract sm from both sides of the equation. Add t to both sides of the equation. Since 2m and sm both contain the variable we are solving for, use the Distributive Property. Divide both sides of the equation by (2 – s).

Solve each equation for the specified variable. Since the problem states to solve for q, move the terms that contain the variable to the same side. Move the other terms to the other side. Move qr to the other side of the equation. Subtract qr from both sides of the equation. Add 18 to both sides of the equation. Since 6q and qr both contain the variable we are solving for, use the Distributive Property. Divide both sides of the equation by (6 – r).

Solve each equation for the specified variable. Since the problem states to solve for x, move the terms that contain the variable to the same side. Move the other terms to the other side. Move xy to the other side of the equation. Add xy to both sides of the equation. Add 2z to both sides of the equation. Since 7x and xy both contain the variable we are solving for, use the Distributive Property. Divide both sides of the equation by (7 + y).

The formula for the volume of a rectangular prism is V = lwh where l is the length, w is the width, and h is the height. Solve the formula for w. Since the problem states to solve for w, divide both sides of the equation by the product of l and h. Find the width of a rectangular prism that has a volume of cubic centimeters, a length of 5.2 centimeters and a height of 4 centimeters. Use the formula from above and plug in what is known. The width of the prism is 3.8 centimeters.

The formula below represents the distance s that a free- falling object will fall near a planet of the Moon in a given time t. In the formula a, represents the acceleration due to gravity. Solve the formula for a. Since the problem states to solve for a, multiply both sides of the equation by 2 and divide both sides by t 2. A free-falling object near the Moon drops 20.5 meters in 5 seconds. What is the value of a for the Moon? Use the formula from above and plug in what is known. The value of a is 1.64 m/s.