Math  /  Algebra

Question13(21y+39)=36\frac{1}{3}(21 y+39)=-36

Studdy Solution

STEP 1

1. The equation 13(21y+39)=36\frac{1}{3}(21y + 39) = -36 is asking us to solve for yy.
2. This is a linear equation involving distribution and basic algebraic operations.
3. The equation can be simplified by eliminating the fraction.

STEP 2

1. Eliminate the fraction by multiplying both sides of the equation by 3.
2. Distribute and simplify the equation.
3. Isolate the variable yy by performing algebraic operations.
4. Check the solution by substituting it back into the original equation.

STEP 3

Multiply both sides of the equation by 3 to eliminate the fraction:
313(21y+39)=3(36) 3 \cdot \frac{1}{3}(21y + 39) = 3 \cdot (-36)
This simplifies to:
21y+39=108 21y + 39 = -108

STEP 4

Now, we need to isolate yy. Start by subtracting 39 from both sides of the equation:
21y+3939=10839 21y + 39 - 39 = -108 - 39
This simplifies to:
21y=147 21y = -147

STEP 5

Next, divide both sides by 21 to solve for yy:
21y21=14721 \frac{21y}{21} = \frac{-147}{21}
This simplifies to:
y=7 y = -7

STEP 6

Check the solution by substituting y=7y = -7 back into the original equation:
13(21(7)+39)=36 \frac{1}{3}(21(-7) + 39) = -36
Simplify inside the parentheses:
21(7)+39=147+39=108 21(-7) + 39 = -147 + 39 = -108
Now, calculate:
13(108)=36 \frac{1}{3}(-108) = -36
Both sides are equal, confirming that the solution is correct.
The solution is:
7 \boxed{-7}

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