Solving Equations

Multiply 2a2/a2+14a+49 * a2-49/2a

1. Factor the numerator for the first rational expression 2a*a 2. Then factor the denominator for the first rational expression (a+7) (a+7) 3. Now, factor the numerator for the second rational expression. (a-7) (a+7) 4. Factor the denominator for the second rational expression (in this case it is already factored) 2a 5. Cancel out 6. Consolidate remainder: a (a-7) /a+7

x2+17/3x=5/3x

1. Find LCD, multiply both sides by LCD a. LCD is 3 b. 3 (x2+17/2x) = 3 (5/3x) c. 3x2+17x = 5 2. Place in standard form for a quadratic equation a. to do this subtract 5x from each side. 3x2+17x = 5x; 3x2+12x 0 3. Factor out the common Monomial factor 3x from the expression 3x2+12x a. 3x2+12x = 0 3x (x+4) = 0 4. Apply the Zero Factor Property a. 3x = 0, or x+4 = 0 5. x = 0, x = -4

7x2-32=0

1. Find the GCF 2. x (7x-32 = 0) 3. Factor the Trinomial 4. According to the Zero Factor Property, if a * b = 0, then a = 0, or b = 0. - In other words, when we multiply two numbers, if the product is zero then one or both of the factors are zero. 5. x = 0, or 7x-32 = 0 6. One root is x= 0, solve the equation to find the second root. 7. 7x =32; Add 32 to both sides 8. Thus, 0 and 32/7 are both roots of this quadratic equation.

14x2-6 = 5x

1. Rewrite in standard form for a quadratic equation - 14x2-5x-6 - place in Descending Order 2. Factor the Trinomial 14x2-5x-6 = 0 3. (7x-6) (2x+1) = 0 4. Apply the Zero Factor Property; set each factor = to zero. - 7x-6 = 0, or 2x+1 = 0 5. Solve for x in these resulting equations to find the roots. 6. x = 6/7, or x = -1/2.

Find the domain of the rational function g (x) = -5x+2/x2+3x-28

1. The domain of a rational function is the set of values that can be used to replace the variable. - Thus, the domain will be all real numbers, EXCEPT those that make the denominator = to zero. 2. Factor the denominator and set them = to zero. 3. The factors that come out of the denominator are your answer.

|5x-4| = 21

1. The solution of an equation of the form |ax+b| = c where a not= 0, and c is a positive number, are those values that satisfy: ax+b = c or, ax+b=-c 2. 5x-4 = 21, or 5x-4 = -21; 5x = 25 (add 5 to both sides) or 5x = -17, divide both sides by 5 3. x=5, or -17/5

Divide 7xy-3x/10x4 div. 7y-3/5x3

1. To divide fractions take the reciprocal of the second fraction and then multiply. 2. 7y-3/5x3 becomes 5x3/7y-3; this is now set as a multiplication problem. 3. Factor out numerators and then denominators into prime factor, factor out common factors. 4.

A = 1/2bh; solve for h

1. To isolate h, divide both sides by the coefficient of h. 2. The coefficient of h is 1/2b 3. Use the reciprocal to multiply 4. 2A/b = h 5. h = 2A/b

P = d+v (n+4); solve for n

1. To solve for n, isolate n on one side of the equation. 2. Begin by removing parentheses: P = d+vn+4v 3. Now to obtain the term containing n on one side of the equation, subtract d from both sides of the equation and subtract 4v from both sides of the equation. P-d = vn +4v P-d-4v =vn 4. Isolate n by dividing both sides of the equation by the coefficient of n P-d-4v/v = n 5. n = P-d-4v/v

Zero Factor Property

If a*0 = 0, then a - 0, or b = 0. - In other words, when we multiply two numbers, if the product is zero, then one or both of the factors are zero. - When this shows up, one of your factors will be x = 0, and you just have to solve for the other factor.

The volume of a rectangular solid can be written as V = LWH, where L is the length of the solid, W is the width and H is the height. A box of cereal has a width of 4 inches. Its height is 2 inches longer than its length. If the volume of the box is 780 cubic inches, what are the length and height of the box?

1. Understand the problem by reading the given information. Then write an equation by using the formula and solve it to find the length of the box. 2. Let x be the length of the box. The the height of the box is x +2. 3. Write the equation of the volume by replacing V (volume) by 780, L (length) by x, W(width) by 4, and H (height) by x+2. So: V = LHW is 780=(x) (4) (x+2) 4. Solve: Divide each side by 4: a. 195 = (x) (x+2) Distribute x on the expression x+2 and simplify. b. 195 = x2+2x Subtract 195 from each side c. 195 = x2+2x is 0 = x2+2x-195 Factor the trinomial d. (x-13) (x+15) = 0 Apply the Zero Factor Property e. x+15 = 0, or x-13 = 0 Simplify f. x = -15, x = 13 5. The length of the box must be a positive number, therefore you can disregard -15. Thus, the length of the box is 13 inches. 6. Find the height by substituting in x =13. a. x+2 = 13+2 = 15 7. Check to see if the height is 2 inches longer then the length a. 15-2 = 13 13=13 (true) 8. Now check if the volume of the box is 780 cubic inches substituting the values found into the V=LWH formula. 9. the length of the box is 13 inches, the height of the box is 15 inches.