Equivalent fractions:  2nd Level

Back to Level 1

Example 1.   Write the missing numerator.

 34 = ___?__8x − 12

Answer.   What times 4 produced 8x − 12?  In other words, we must expect that 4, the denominator on the left, will be a factor on the right.

 34 = ___?__8x − 12 = ___?___4(2x − 3)

The denominator 4 has been multiplied by 2x − 3; therefore, the numerator 3 also must be multiplied by 2x − 3.

 34 = 3(2x − 3)4(2x − 3) = 6x − 9 8x − 12

Example 2.   Write the missing numerator.

 __5_x − 3 = ____?____x2 − 4x + 3

Answer.   What times  x − 3  produced  x2 − 4x + 3?   For, we must expect that x − 3 will be a factor of x2 − 4x + 3.

 __5_x − 3 = ____?____x2 − 4x + 3 = _____?_____(x − 3)(x − 1)

x − 3 has been multiplied by x − 1; therefore, 5 also must be multiplied by x − 1:

 __5_x − 3 = __5(x − 1)__(x − 3)(x − 1) = __5x − 5__x2 − 4x + 3

Problem 1.   Factor the new denominator, and write the missing numerator.

 a) 25 = ___?___15x + 20 = ___?___5(3x + 4) = 2(3x + 4)5(3x + 4) = _6x + 8_15x + 20
 b) 2x = __?__x2 − x = __  ? __x(x − 1) = 2(x − 1)x(x − 1) = 2x − 2x2 − x
 c) _5_2x2 = ____?___4x3 + 6x2 = __ __?_ __2x2(2x + 3) = 5(2x + 3)2x2(2x + 3) = 10x + 154x3 + 6x2
 d) 4x = ___?___8x2 − 3x = _ __?___x(8x − 3) = 4(8x − 3)x(8x − 3) = 32x − 128x2 − 3x
 e) _2_3x4 = ____?___6x6 + 3x4 = ___ _?_ __3x4(2x2 + 1) = _2(2x2 + 1)_3x4(2x2 + 1) = _4x2 + 2_6x6 + 3x4

The following problems assumes skill with Quadratic Trinomials,  Perfect Square Trinomials, and the Difference of Two Squares.

Problem 2.   Factor the new denominator, and write the missing numerator.

 a) __2__x + 4 = ____?____x2 + 6x + 8 = __ 2(x + 2)__(x + 4)(x + 2) = __2x + 4__x2 + 6x + 8
 b) x + 3x − 2 = ____?____x2 + x − 6 = __ (x + 3)2__(x − 2)(x + 3) = x2 + 6x + 9x2 + x − 6
 c) __1__x − 2 = __?__x2 − 4 = ___ x + 2___(x − 2)(x + 2) = x + 2x2 − 4
 d) __x__x + 2 = ____?____x2 + 5x + 6 = __x(x + 3)__(x + 2)(x + 3) = __x2 + 3x_x2 + 5x + 6
 e) x + 3x − 1 = ____?____x2 − 4x + 3 = (x + 3)(x − 3)(x − 1)(x − 3) = __x2 − 9__x2 − 4x + 3
 f) x − 5x + 5 = ___?__x2 − 25 = __ (x − 5)2__(x + 5)(x − 5) = x2 − 10x + 25    x2 − 25
 g) 2x − 1x − 5 = ____?____2x2 − 9x − 5 = (2x − 1)(2x + 1) (x − 5)(2x + 1) = __4x2 − 1__2x2 − 9x − 5

Problem 3.   Reduce to lowest terms. See Lesson 19, Problems 10, 12, 14.

 a) x5 − 1 x − 1 = x4 + x3 + x2 + x + 1

x − 1 is a factor of x5 − 1.

 b) x4 − 1 x − 1 = x3 + x2 + x + 1
 c) x3 − 1 x − 1 = x2 + x + 1
 d) x7 − 1 x − 1 = x6 + x5 + x4 + x3 + x2 + x + 1

The following problem depends on knowing that  ba  is the negative of ab.  (Lesson 7.)

Problem 4.   Simplify.

 a) 2 − xx − 2 = −1 b) x(x − 3)  3 − x = x(−1) = −x
 c) __1 − x_x2(x − 1) = 1 x2 (−1) = − 1 x2
 d) 3x − 6 2 − x = 3(x − 2)  2 − x = 3(−1) = −3
 e) x2 − 7x + 12    4 −x = (x − 4)(x − 3)     4 −x = −1(x − 3) = 3 − x
 f) __1 − x__ x2 − 2x + 1 = ___1 − x__  (x − 1)(x − 1) = −1  x − 1 = 1     −(x − 1) = 1   1 − x

Next Lesson:  Negative exponents

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