Answer:
13
Step-by-step explanation:
We can multiply both the whole equation by [tex]c^2-7c-18[/tex]. This is the LCM of all of the denominators. Notice that when [tex]c^2-7c-18[/tex] is factored, we obtain [tex](c - 9)(c+2)[/tex].
Step 1: Multiply Left Side by [tex]c^2-7c-18[/tex]
[tex]\frac{5}{c-9}\times(c^2-7c-18)=\frac{5}{c-9}\times(c-9)(c+2)=5(c+2)=5(c)+5(2)=5c+10[/tex]
Step 2: Multiply Right Side by [tex]c^2-7c-18[/tex]
An easier approach is to multiply each individual fraction like so:
[tex]\frac{5c-2}{c^2-7c-18}\times(c^2-7c-18)=5c-2[/tex]
[tex]\frac{3}{c+2}\times(c^2-7c-18)=\frac{3}{c+2}\times(c-9)(c+2)=3(c-9)=3c-27[/tex]
Then, we add them:
[tex]5c-2+(3c-27)=5c-2+3c-27=8c-29[/tex]
Therefore the whole equation is now:
[tex]5c+10=8c-29[/tex]
Step 3: Solve for "c"
Subtract 5c from both sides:
[tex]10=3c-29[/tex]
Add 29 to both sides:
[tex]39=3c[/tex]
Divide both sides by 3
[tex]c=13[/tex]
Step 4: Plug 13 in for "c" to check our work
[tex]\frac{5}{13-9}=\frac{5(13)-2}{169-91-14}+\frac{3}{13+2}\\\\\frac{5}{4}=\frac{63}{60}+\frac{3}{15}\\\\\frac{5}{4}=\frac{21}{20}+\frac{1}{5}\\\\\frac{5}{4}=\frac{21}{20}+\frac{1}{5}\\\\\frac{5}{4}=\frac{21}{20}+\frac{4}{20}\\\\\frac{5}{4}=\frac{25}{20}\\\\\frac{5}{4}=\frac{5}{4}\Rightarrow Correct![/tex]
Therefore,
c = 13
