University:
California State University, Northridge
Course:
MATH 320 | Foundations of Higher Mathematics
Academic year:

2024
Views:

308

Pages:

13
Author:

Ciara Meyer

§ 14.5 Solving Equations Containing Rational Expressions Solving Equations First note that an equation contains an equal sign and an expression does not. To solve EQUATIONS containing rational expressions, clear the fractions by multiplying both sides of the equation by the LCD of all the fractions. Then solve as in previous sections. Note: this works for equations only, not simplifying expressions. 45 Solving Equations Example Solve the following rational equation. 5 7 +1 = 3x 6  7  5 6 x + 1 =  6 x  3x   6  10 + 6 x = 7 x 10 = x Check in the original equation. 5 7 +1 = 3 ⋅10 6 5 7 +1 = 30 6 7 1 +1 = 6 6 true 46 Solving Equations Example Solve the following rational equation. 1 1 1 − = 2 2 x x + 1 3x + 3x  1   1  1 6 x( x + 1) 6 x( x + 1) −  =   2 x x + 1   3 x( x + 1)  3( x + 1) − 6 x = 2 3x + 3 − 6 x = 2 3 − 3x = 2 − 3 x = −1 x= 1 3 Continued. 47 Solving Equations Example Continued Substitute the value for x into the original equation, to check the solution. 1 1 1 − = 2 1 1 2 +1 3 1 + 3 1 3 3 3 3 ( ) ( ) ( ) ( ) 3 3 1 − = 2 4 1 +1 3 6 3 3 − = 4 4 4 true So the solution is x = 13 48 Solving Equations Example Solve the following rational equation. x+2 1 1 = − 2 x + 7 x + 10 3 x + 6 x + 5 1  x+2   1  3( x + 2 )( x + 5) 2 − 3( x + 2 )( x + 5) =  x + 7 x + 10   3 x + 6 x + 5  3( x + 2 ) = ( x + 5) − 3( x + 2 ) 3x + 6 = x + 5 − 3x − 6 3x − x + 3x = 5 − 6 − 6 5 x = −7 x = −7 5 Continued. 49 Solving Equations Example Continued Substitute the value for x into the original equation, to check the solution. ( −7 +2 1 1 5= − 2 7 7 −7 + 7 − 7 + 10 3 − 5 + 6 − 5 + 5 5 5 3 1 1 5 = − 49 − 49 + 10 − 21 + 6 18 5 25 5 5 5 =5− 5 18 9 18 true So the solution is x = − 7 5 ) ( ) ( ) ( ) 50 Solving Equations Example Solve the following rational equation. 1 2 = x −1 x +1  1   2  (x − 1)(x + 1) ( x − 1)( x + 1) =  x −1   x +1  x + 1 = 2( x − 1) x +1 = 2x − 2 3= x Continued. 51 Solving Equations Example Continued Substitute the value for x into the original equation, to check the solution. 1 2 = 3 −1 3 +1 1 2 = 2 4 true So the solution is x = 3. 52 Solving Equations Example Solve the following rational equation. 12 3 2 + = 2 9−a 3+ a 3− a 3   2   12 (3 − a )(3 + a ) + (3 − a )(3 + a ) = 2 3+ a   3− a  9−a 12 + 3(3 − a ) = 2(3 + a ) 12 + 9 − 3a = 6 + 2a 21 − 3a = 6 + 2a 15 = 5a 3=a Continued. 53 Solving Equations Example Continued Substitute the value for x into the original equation, to check the solution. 12 3 2 2 + 3+3 = 3−3 9−3 12 3 2 + = 0 5 0 Since substituting the suggested value of a into the equation produced undefined expressions, the solution is ∅. 54 Solving Equations with Multiple Variables Solving an Equation With Multiple Variables for One of the Variables 1) Multiply to clear fractions. 2) Use distributive property to remove grouping symbols. 3) Combine like terms to simplify each side. 4) Get all terms containing the specified variable on the same side of the equation, other terms on the opposite side. 5) Isolate the specified variable. 55 Solving Equations with Multiple Variables Example Solve the following equation for R1 1 1 1 = + R R1 R2 1  1  1 RR1 R2   =  +  RR1 R2  R   R1 R2  R1 R2 = RR2 + RR1 R1 R2 − RR1 = RR2 R1 (R2 − R ) = RR2 RR2 R1 = R2 − R 56