Answer:
a = 4 and b = -2
Step-by-step explanation:
Although the problem stated is formulated a bit strangely, here is what I interpreted the problem as:
[tex]\frac{a}{\sqrt{3}+1}+ \frac{b}{\sqrt{3}-1}= \sqrt{3}-3[/tex]
So, we have to solve for a and b, but the denominators are different. We have to get the same denominators for both fractions, so we cross multiply:
[tex](\frac{a}{\sqrt{3}+1})(\frac{\sqrt{3}-1}{\sqrt{3}-1}) + \frac{b}{\sqrt{3}-1}(\frac{\sqrt{3}+1}{\sqrt{3}+1}) = \sqrt{3}-3[/tex]
[tex]\frac{a(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} + \frac{b(\sqrt{3}+1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \sqrt{3}-3[/tex]
∴ [tex]\frac{a(\sqrt{3}-1)+b(\sqrt{3}+1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \sqrt{3}-3[/tex]
Now, we can simplify the equation a bit. We know that:
[tex](\sqrt{3}+1)(\sqrt{3}-1)=\sqrt{9}-1=2[/tex]
Therefore:
[tex]\frac{a(\sqrt{3}-1)+b(\sqrt{3}+1)}{2} = \sqrt{3}-3[/tex]
[tex]a(\sqrt{3}-1)+b(\sqrt{3}+1) = (2)(\sqrt{3}-3)[/tex]
[tex]a(\sqrt{3}-1)+b(\sqrt{3}+1) = 2\sqrt{3}-6[/tex]
Using logical deduction, we can see that a = 4 and b = -2, as seen here:
[tex](4)(\sqrt{3}-1)+(-2)(\sqrt{3}+1) = 2\sqrt{3}-6[/tex]
[tex]4\sqrt{3}-4-2\sqrt{3}-2 = 2\sqrt{3}-6[/tex]
Combine like terms:
[tex]2\sqrt{3}-6= 2\sqrt{3}-6[/tex]
[tex]\sqrt{3}-3= \sqrt{3}-3[/tex]
