Solve by Factoring 16y+3y^2=35

$16 y + 3 y^{2} = 35$

Step 1

Subtract $35$ from both sides of the equation.

$16 y + 3 y^{2} - 35 = 0$

Step 2

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

$16 u + 3 u^{2} - 35 = 0$

Step 3

Factor by grouping.

Reorder terms.

$3 u^{2} + 16 u - 35 = 0$

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot - 35 = - 105$ and whose sum is $b = 16$ .

Factor $16$ out of $16 u$ .

$3 u^{2} + 16 (u) - 35 = 0$

Rewrite $16$ as $- 5$ plus $21$

$3 u^{2} + (- 5 + 21) u - 35 = 0$

Apply the distributive property.

$3 u^{2} - 5 u + 21 u - 35 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 u^{2} - 5 u) + 21 u - 35 = 0$

Factor out the greatest common factor (GCF) from each group.

$u (3 u - 5) + 7 (3 u - 5) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 u - 5$ .

$(3 u - 5) (u + 7) = 0$

Step 4

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

$(3 y - 5) (y + 7) = 0$

Step 5

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

$3 y - 5 = 0$

$y + 7 = 0$

Step 6

Set $3 y - 5$ equal to $0$ and solve for $y$ .

Set $3 y - 5$ equal to $0$ .

$3 y - 5 = 0$

Solve $3 y - 5 = 0$ for $y$ .

Add $5$ to both sides of the equation.

$3 y = 5$

Divide each term in $3 y = 5$ by $3$ and simplify.

Divide each term in $3 y = 5$ by $3$ .

$\frac{3 y}{3} = \frac{5}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{5}{3}$

Divide $y$ by $1$ .

$y = \frac{5}{3}$

Step 7

Set $y + 7$ equal to $0$ and solve for $y$ .

Set $y + 7$ equal to $0$ .

$y + 7 = 0$

Subtract $7$ from both sides of the equation.

$y = - 7$

Step 8

The final solution is all the values that make $(3 y - 5) (y + 7) = 0$ true.

$y = \frac{5}{3}, - 7$

Solve by Factoring 16y+3y^2=35

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