How to solve 7y+6-2y+3+3y-4

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

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Question

$$7y+6-2y+3+3y-4$$

Evaluate

$8y+5$

Solution Steps

Combine $7y$ and $-2y$ to get $5y$.

$$5y+6+3+3y-4$$

Add $6$ and $3$ to get $9$.

$$5y+9+3y-4$$

Combine $5y$ and $3y$ to get $8y$.

$$8y+9-4$$

Subtract $4$ from $9$ to get $5$.

$$8y+5$$

Show Solution

Differentiate w.r.t. y

$8$

Steps Using Derivative Rule for Sum

Combine $7y$ and $-2y$ to get $5y$.

$$\frac{\mathrm{d}}{\mathrm{d}y}(5y+6+3+3y-4)$$

Add $6$ and $3$ to get $9$.

$$\frac{\mathrm{d}}{\mathrm{d}y}(5y+9+3y-4)$$

Combine $5y$ and $3y$ to get $8y$.

$$\frac{\mathrm{d}}{\mathrm{d}y}(8y+9-4)$$

Subtract $4$ from $9$ to get $5$.

$$\frac{\mathrm{d}}{\mathrm{d}y}(8y+5)$$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is $0$. The derivative of $ax^{n}$ is $nax^{n-1}$.

$$8y^{1-1}$$

Subtract $1$ from $1$.

$$8y^{0}$$

For any term $t$ except $0$, $t^{0}=1$.

$$8\times 1$$

For any term $t$, $t\times 1=t$ and $1t=t$.

$$8$$