Math  /  Algebra

Question3=c31c=\begin{array}{l}3=\frac{c}{3}-1 \\ c=\square\end{array}

Studdy Solution

STEP 1

What is this asking? We need to find the value of cc that makes the equation 3=c313 = \frac{c}{3} - 1 true. Watch out! Don't forget to follow the order of operations and keep track of the negative signs.

STEP 2

1. Isolate the term with cc.
2. Solve for cc.

STEP 3

We want to get c3\frac{c}{3} by itself.
The equation says that c3\frac{c}{3} minus 11 equals 33.
So, to isolate the term with cc, we need to add 11 to both sides of the equation.
This is like a seesaw – to keep it balanced, what we do to one side, we *must* do to the other!
3+1=c31+13 + 1 = \frac{c}{3} - 1 + 1

STEP 4

Now, let's simplify!
On the left side, 3+13 + 1 gives us **44**.
On the right side, 1+1-1 + 1 adds to zero, leaving us with just c3\frac{c}{3}.
4=c34 = \frac{c}{3}

STEP 5

We're *so* close!
We have c3=4\frac{c}{3} = 4.
To get cc all by itself, we need to multiply both sides of the equation by **33**.
Remember, it's like a seesaw, gotta keep it balanced!
Multiplying c3\frac{c}{3} by 33 gives us 3c3\frac{3 \cdot c}{3}.
34=3c33 \cdot 4 = 3 \cdot \frac{c}{3}

STEP 6

On the left side, 343 \cdot 4 gives us **1212**.
On the right side, the 33 in the numerator and the 33 in the denominator divide to one, leaving us with just cc.
12=c12 = c

STEP 7

We found that c=12c = 12.
Let's check our work!
Substituting c=12c = 12 back into the original equation gives us 3=12313 = \frac{12}{3} - 1, which simplifies to 3=413 = 4 - 1, and finally 3=33 = 3.
It works!
So, c=12c = \boxed{12}.

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