Question Solve the equation $9 c(c-11)+10(5 c-3)=3 c(c+5)+c(6 c-3)-30$ for $c$ .

Studdy Solution

STEP 1

Assumptions
1. We are given a quadratic equation in the form of: $9 c(c-11)+10(5 c-3)=3 c(c+5)+c(6 c-3)-30$
2. We need to solve for the variable $c$ .

STEP 2

Expand both sides of the equation to get rid of the parentheses. This will help us to combine like terms.
$9c^2 - 99c + 50c - 30 = 3c^2 + 15c + 6c^2 - 3c - 30$

STEP 3

Combine like terms on both sides of the equation.
On the left side: $9c^2 - 99c + 50c - 30 = 9c^2 - 49c - 30$
On the right side: $3c^2 + 15c + 6c^2 - 3c - 30 = 9c^2 + 12c - 30$

STEP 4

Now we have simplified equations on both sides:
$9c^2 - 49c - 30 = 9c^2 + 12c - 30$

STEP 5

Since we have $9c^2$ and $-30$ on both sides of the equation, we can subtract $9c^2$ and add $30$ to both sides to eliminate those terms.
$9c^2 - 49c - 30 - 9c^2 + 30 = 9c^2 + 12c - 30 - 9c^2 + 30$

STEP 6

Simplify the equation after eliminating the terms.
$-49c = 12c$

STEP 7

Add $49c$ to both sides to isolate the variable $c$ on one side of the equation.
$-49c + 49c = 12c + 49c$

STEP 8

Combine like terms to solve for $c$ .
$0 = 61c$

STEP 9

Divide both sides by $61$ to solve for $c$ .
$\frac{0}{61} = \frac{61c}{61}$

STEP 10

Simplify the equation to find the value of $c$ .
$0 = c$
The solution to the equation is: $c = 0$

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