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Sagot :

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

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