Module 3 topic 3 notes
- 1. Module 3, Topic 3 Notes Solving Literal Equations
- 2. Solving Literal Equations… Means we are solving equations for a specific variable, but won’t get a numerical answer like before. Instead of getting x = 2, we might get x = 6 + 5y. Means “getting a letter by itself”. If we are solving an equation for y, we want the equation to say y = when we are done. Still means we use inverse operations!
- 3. Examples of Solving Literal Equations Solve I = prt for r. This means we want to have an equation that says r =. To get r by itself, we would divide by p and t. I = prt pt pt r = I pt Solve 3x + 2y = -16 for y. This means we want to get y by itself. 3x + 2y = -16 -3x -3x 2y = -3x – 16 2 2 2 y = -3/2x – 8
- 4. More Examples Solve 5x – 20y = -40 for x. This means we want to get x by itself. 5x – 20y = -40 +20y +20y 5x = 20y – 40 5 5 5 x = 4y – 8 Solve -3x – 8y = -16 for y. This means we want to get y by itself. -3x – 8y = -16 +3x +3x -8y = 3x – 16 -8 -8 -8 y = -3/8x + 2
- 5. More About Literal Equations Being able to “rearrange” equations is very important throughout Algebra I and other math classes. You will have to be able to solve for y in our slope modules, because y = mx + b is slope-intercept form. You will have to be able to solve formulas for given variables in Geometry so that you can correctly find the answer to a problem.
- 6. Examples of Problems that Use Literal Equations Direct variation can be modeled by the equation y = kx, where k is a constant (a regular number). Jose is making cupcakes, and the number of eggs needed varies directly as the cups of flour needed. If 5 cups of flour are needed when using 8 eggs, find k (the constant of variation), if x represents the number of eggs and y represents the cups of flour. First, solve y = kx for k (because that’s what you want to know). x x y/x = k Then, plug in 5 for y (flour) and 8 for x (eggs). 5/8 = k Done!
- 7. Another Example Bricklayers use the formula N = 7LH to estimate the number of bricks N needed to build a wall of length L and height H. What is the height of a wall that is 30 feet long and requires 2310 bricks to build? First, solve your equation for H, since height is what we are trying to find. N = 7LH 7L 7L H = N 7L Then, plug in 2310 for N (# of bricks) and 30 for L (length). H = 2310 = 2310 = 11 7(30) 210 The height of the wall is 11 feet.
- 8. Some Extra Practice for You! Here’s a website with some examples you can use to check your understanding. Click the drop-down menus that say “Answer” and you can get helpful hints: http://www.regentsprep.org/Regents/math/ALGEBRA/AE4/litPrac.htm