Hepl solve

Question

2v^2-9v+4/2v^2+3v-2 / 8v^2-6v+1/4v^2+7v-2

Answer 1

Troy, first let's clear up the equation/question.  Using parentheses and brackets, it looks to me that you have been given the following

[(2v²-9v+4)/(2v²+3v-2)]/[(8v²-6v+1)/(4v²+7v-2)]

In this situation you are not "solving" this, but you are being asked to simplify it.  This usually requires factoring each of the 4 quadratic expressions.  And once factored, cancel out or manipulate like terms.  So let's look at each of these expressions in turn.

(2v²-9v+4)

To factor this, since the last sign is + (+4) and the second sign (-9) is minus the factors have to both be subtractions, i.e. (__v-__)(__v-__).  The first blank has to multiply to 2, so the options are 1 or 2.  The second blank has to multiply to 4, so the options are 1, 2 or 4.  Whichever terms you choose also have to produce -9 as the second term.  Looking at the options 2x4 =8 and 1x1 = 1, and 8+1=9.  So the factoring should look like this (2v-1)(v-4)=(2v²-9v+4)

(2v²+3v-2)

Here the last sign is "-", so one factor has to be (__v+___) and the other has to be (__v-__).  This time the options are to fill in the blanks are 1 or 2.  Because we want +3v, we probably want to have 2x2=4 and -1x1=-1, so that 4-1=3. So the factoring looks like this (2v-1)(v+2)=(2v²+3v-2).  Notice that 2v-1 is a term in both the first and second quadratic.

(8v²-6v+1)

Like the first quadratic, we have a plus at the end and a - second term so the factoring has to be
(__v-__)(__v-__).  1 factors to 1x1, the options for 8 are 1 & 8 or 2 & 4. Since 1+8=9 and 2+4=6, 2+4 will produce the desired middle term, so (2v-1)(4v-1)=(8v²-6v+1).  Again 2v-1 appears!

(4v²+7v-2)

Like the second quadratic, we have a minus at the end, so one factor has to be (__v+___) and the other (__v-___).  The options for 2 are 1 and 2 and for 4 are 2x2 or 4x1.  Since we need a plus 7, we probably want 4x2=8 and 1x-1=-1 since 8-1=7. So, (4v-1)(v+2)=(4v²+7v-2)

With these factors, you can substitute them into the original equation, and I've placed the factors into the ()'s that the quadratics were originally in.

[((2v-1)(v-4))/((2v-1)(v+2))]/[((2v-1)(4v-1))/((4v-1)(v+2))]

Now this is ugly looking.  When you have a fraction over a fraction [e.g. (1/9)/(2/7)], I recommend multiplying the equation by the the reciprocal of the denominator over the reciprocal of the denominator (bottom fraction), e.g. [(1/9)/(2/7)]x[(7/2)/(7/2)]=7/18.  This does not change the value of the expression because x/x=1, so we are just multiplying by one, but it eliminates the bottom fraction because multiplying a number by its reciprocal makes it equal to 1. Here this becomes

[((2v-1)(v-4))/((2v-1)(v+1))]x[((4v-1)(v+2))/((2v-1)(4v-1))]

Now, before you multiply, see if you can cancel any terms.  For example 2v-1 appears in the numerator and denominator of the first term, so they can cancel out.  Similarly 4v-1 appears in both the numerator and the denominator of the second term, so they can cancel out.

[((2v-1)(v-4))/((2v-1)(v+1))]x[((4v-1)(v+2))/((2v-1)(4v-1))]

Now you can combine the terms, i.e.

[(v-4)(v+2)]/[(v+2)(2v-1)], and then you can cancel the v+2 term in the numerator and denominator, so the answer is (v-4)/(2v-1).

I hope this helps.