ACT Math 2.1 Intermediate Algebra
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ACT Math Intermediate Algebra Drill 2, Problem 1. Solve the equation.
|ACT Math||Intermediate Algebra|
|ACT Mathematics||Intermediate Algebra|
Interpret the structure of expressions
Write expressions in equivalent forms to solve problems
|Elementary Algebra||Solve quadratic equations by factoring|
|Foreign Language||Arabic Subtitled|
|Intermediate Algebra||Fractional equations|
Radical and rational expressions
|Product Type||ACT Math|
One: multiplying by the inverse -- so we have to flip the 2nd fraction around and make
it times rather than divided by...
And then two: We clearly have a factorable expression here.
For the first one we have an a in each of these parts of the parentheses...
...then we have to figure out what 2 numbers multiply together to get negative 12 and also
add together to get negative 1.
What are the factors of 12? 1 and 12? Yeah... but that won't work.
2 and 6 don't work either, because when added, no matter which is positive and which
is negative, they won't total "1."
3 and 4 however... DO work.
So we want the ADDED part to be a negative 1... negative 4 plus 3 is negative 1, and
negative 4 times 3 is negative 12, so... we're good.
The first part of our answer is quantity a plus 3 times quantity a minus 4.
Now to the denominator.
We need 2 numbers that when multiplied are 2... and when added are 3.
Well... 2 only has 2 and 1 as factors... and luckily, if we add 2 and 1 we get 3. So the
bottom is a quick calculation.
It's quantity a plus 2 times quantity a plus 1.
Now we just rewrite everything so it's fully factored and... cross-out-able.
The quantity a plus 3 over the quantity a plus 3 becomes 1; we can trash it.
Same deal with the quantity a plus 2.
And we are left with just quantity a minus 4 over a plus 1...
...and the answer is C.