Answer:
[tex]a=3[/tex] and [tex]b=5[/tex].
Step-by-step explanation:
So I believe the problem is this:
[tex]\frac{2(4x+3)}{x-3}(x+7)}=\frac{a}{x-3}+\frac{b}{x+7}[/tex]
where we are asked to find values for [tex]a[/tex] and [tex]b[/tex] such that the equation holds for any [tex]x[/tex] in the equation's domain.
So I'm actually going to get rid of any domain restrictions by multiplying both sides by (x-3)(x+7).
In other words this will clear the fractions.
[tex]\frac{2(4x+3)}{x-3}(x+7)}\cdot(x-3)(x+7)=\frac{a}{x-3}\cdot(x-3)(x+7)+\frac{b}{x+7}(x-3)(x+7)[/tex]
[tex]2(4x+3)=a(x+7)+b(x-3)[/tex]
As you can see there was some cancellation.
I'm going to plug in -7 for x because x+7 becomes 0 then.
[tex]2(4\cdot -7+3)=a(-7+7)+b(-7-3)[/tex]
[tex]2(-28+3)=a(0)+b(-10)[/tex]
[tex]2(-25)=0-10b[/tex]
[tex]-50=-10b[/tex]
Divide both sides by -10:
[tex]\frac{-50}{-10}=b[/tex]
[tex]5=b[/tex]
Now we have:
[tex]2(4x+3)=a(x+7)+b(x-3)[/tex] with [tex]b=5[/tex]
I notice that x-3 is 0 when x=3. So I'm going to replace x with 3.
[tex]2(4\cdot 3+3)=a(3+7)+b(3-3)[/tex]
[tex]2(12+3)=a(10)+b(0)[/tex]
[tex]2(15)=10a+0[/tex]
[tex]30=10a[/tex]
Divide both sides by 10:
[tex]\frac{30}{10}=a[/tex]
[tex]3=a[/tex]
So [tex]a=3[/tex] and [tex]b=5[/tex].
