Algebra Examples

${(y - 1)}^{2} - 13 (y - 1) + 36 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Let $u = y - 1$ . Substitute $u$ for all occurrences of $y - 1$ .

$u^{2} - 13 u + 36 = 0$

Step 1.2

Factor $u^{2} - 13 u + 36$ using the AC method.

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Step 1.2.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $36$ and whose sum is $- 13$ .

$- 9, - 4$

Step 1.2.2

Write the factored form using these integers.

$(u - 9) (u - 4) = 0$

Step 1.3

Replace all occurrences of $u$ with $y - 1$ .

$(y - 1 - 9) (y - 1 - 4) = 0$

Step 2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$y - 1 - 9 = 0$

$y - 1 - 4 = 0$

Step 3

Set $y - 1 - 9$ equal to $0$ and solve for $y$ .

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Step 3.1

Set $y - 1 - 9$ equal to $0$ .

$y - 1 - 9 = 0$

Step 3.2

Solve $y - 1 - 9 = 0$ for $y$ .

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Step 3.2.1

Subtract $9$ from $- 1$ .

$y - 10 = 0$

Step 3.2.2

Add $10$ to both sides of the equation.

$y = 10$

Step 4

Set $y - 1 - 4$ equal to $0$ and solve for $y$ .

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Step 4.1

Set $y - 1 - 4$ equal to $0$ .

$y - 1 - 4 = 0$

Step 4.2

Solve $y - 1 - 4 = 0$ for $y$ .

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Step 4.2.1

Subtract $4$ from $- 1$ .

$y - 5 = 0$

Step 4.2.2

Add $5$ to both sides of the equation.

$y = 5$

Step 5

The final solution is all the values that make $(y - 1 - 9) (y - 1 - 4) = 0$ true.

$y = 10, 5$