Answer:
a=6, b=2, c=6, d=2, e=6, f=6, g=1
Explanation:
You want the values of letters a–g that show the correct simplification of ...
![(2)/(x^2-36)-(1)/(x^2+6x)=(2)/((x+6)(x-6))-(1)/(x(x+a))\\\\=(bx)/((x+6)(x-6)x)-(x-c)/((x+6)(x-6)x)=(dx-x+e)/((x+6)(x-6)x)\\\\=(x+f)/((x+6)(x-6)x)=(g)/(x(x+6))](https://img.qammunity.org/2023/formulas/mathematics/college/2yy1kl5nv4v3wraiixnnudpk896dyftnno.png)
Solution
The difference of squares is factored as (a² -b²) = (a +b)(a -b), so the denominator of the first term can be factored. The common factor x can be factored out of the denominator of the second term. This gives ...
![(2)/(x^2-36)-(1)/(x^2+6x)= (2)/((x+6)(x-6))-(1)/(x(x+6))\quad a=6](https://img.qammunity.org/2023/formulas/mathematics/college/7n60s6nft0x83mokvnmkl44r4pe04yeta6.png)
Multiplying the first term by x/x and the second term by (x-6)/(x-6) gives ...
![(2x)/((x+6)(x-6)x)-(x-6)/((x+6)(x-6)x)\quad b=2,\ c=6](https://img.qammunity.org/2023/formulas/mathematics/college/x1r93kyc580mdh5yev1zwlpv32uc2ctr9o.png)
Writing the sum over the common denominator, we have ...
![(2x-x+6)/((x+6)(x-6)x)\quad d=2,\ e=6](https://img.qammunity.org/2023/formulas/mathematics/college/wq3nteq5u45v0cpahc6ui473jgi7f7asyk.png)
Collecting terms in the numerator gives ...
![(x+6)/((x+6)(x-6)x)\quad f=6](https://img.qammunity.org/2023/formulas/mathematics/college/abieep97px8uzhdg59okdyxjwbvgapfu5q.png)
Finally, cancelling the common factor (x+6) from numerator and denominator, we have the simplified form ...
![(1)/(x(x+6))\quad g=1](https://img.qammunity.org/2023/formulas/mathematics/college/q60v3q5x12nxxqqvybpi6fiqy3diptlkte.png)
The values of the letters are ...
a=6, b=2, c=6, d=2, e=6, f=6, g=1