QuestionSolve the equation and check. $y=\square(y-3(y-2)=6(y+1)$

Studdy Solution

STEP 1

1. The given equation is linear and can be solved using basic algebraic operations.
2. The equation involves distributing, combining like terms, and isolating the variable $y$ .

STEP 2

1. Distribute the constants inside the parentheses.
2. Combine like terms on both sides of the equation.
3. Isolate the variable $y$ by performing algebraic operations.
4. Check the solution by substituting it back into the original equation.

STEP 3

Distribute the constants inside the parentheses on both sides of the equation.
$7y - 3(y - 2) = 6(y + 1)$
$7y - 3y + 6 = 6y + 6$

STEP 4

Combine like terms on both sides of the equation.
$(7y - 3y) + 6 = 6y + 6$
$4y + 6 = 6y + 6$

STEP 5

Isolate the variable $y$ by performing algebraic operations. First, subtract $6y$ from both sides.
$4y + 6 - 6y = 6y + 6 - 6y$
$4y - 6y + 6 = 6$
$-2y + 6 = 6$

STEP 6

Continue isolating $y$ by subtracting 6 from both sides.
$-2y + 6 - 6 = 6 - 6$
$-2y = 0$

STEP 7

Solve for $y$ by dividing both sides by $-2$ .
$y = \frac{0}{-2}$
$y = 0$

STEP 8

Check the solution by substituting $y = 0$ back into the original equation.
Original equation:
$7y - 3(y - 2) = 6(y + 1)$
Substitute $y = 0$ :
$7(0) - 3(0 - 2) = 6(0 + 1)$
$0 - 3(-2) = 6(1)$
$0 + 6 = 6$
$6 = 6$
The solution satisfies the original equation.
The solution to the equation is $y = 0$ .

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