Lori C. answered 03/17/18
Tutor
5.0
(21)
Algebra, Trigonometry and Calculus
This software doesn't display fractions well. So rewrite this on your paper using horizontal lines for the fractions.
Notice on the left hand side, the denominators are (x - 5) and (x -2). In order to add them, you must have a common denominator.
The easy way to do that is to multiply
A/(x - 5) by the fraction (x - 2)/(x - 2). That makes that first fraction be
A( x - 2)/[(x- 5)(x - 2)].
Next, multiply B/(x - 2) by (x - 5)/(x - 5). This fraction becomes
B(x - 5)/[(x - 2)(x - 5)]
Now we can add those 2 fractions because they have the same denominator. We get...
[A( x - 2) + B( x - 5)]/[(x - 2)(x - 5)]
Remember that this is the left hand side. We have not done anything to the right hand side yet.
Let's multiply out that big fraction ...
(Ax - 2A + Bx - 5B)/(x2 - 7x + 10)
Now look at the right hand side. Notice that the denominator on the right is the same as the denominator we multiplied out. That is what we want to see!
Since the 2 expressions are equal AND the denominators are equal, the numerators must be equal as well! We can totally ignore the denominators until the final check!
Just set the numerators equal and solve. You should see...
Ax - 2A +Bx - 5B = 10x - 41.
Regroup to get
(Ax + Bx) - (2A + 5B) = 10x - 41
So (A + B) x = 10 x
And 2A + 5B = 41
You now have 2 equations and 2 unknowns (A and B)...... Let's use substitution.
A + B = 10 is the same as A = 10 - B.
Put that in for A in the second equation...
2A + 5B = 41
2(10-B) + 5B = 41
20 - 2B + 5B= 41
20 +3B = 41
3B = 21 and B = 7
Now substitute for B back in the first equation....
A + B = 10
A + 7 = 10 and A = 3
Let's check to be sure..
3 + 7 = 10.....check
2(3) + 5(7) = 6 + 35 = 41 .....check.
A = 3 and B = 7
You should substitute back into the original problem...just to make sure. But my battery is dead!
I hope this helps!