How to solve 9z+3-4z+2+2z+5

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.

Solve 9z+3-4z+2+2z+5 | Equatix

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Question

$$9z+3-4z+2+2z+5$$

Evaluate

$7z+10$

Solution Steps

Combine $9z$ and $-4z$ to get $5z$.

$$5z+3+2+2z+5$$

Add $3$ and $2$ to get $5$.

$$5z+5+2z+5$$

Combine $5z$ and $2z$ to get $7z$.

$$7z+5+5$$

Add $5$ and $5$ to get $10$.

$$7z+10$$

Show Solution

Differentiate w.r.t. z

$7$

Steps Using Derivative Rule for Sum

Combine $9z$ and $-4z$ to get $5z$.

$$\frac{\mathrm{d}}{\mathrm{d}z}(5z+3+2+2z+5)$$

Add $3$ and $2$ to get $5$.

$$\frac{\mathrm{d}}{\mathrm{d}z}(5z+5+2z+5)$$

Combine $5z$ and $2z$ to get $7z$.

$$\frac{\mathrm{d}}{\mathrm{d}z}(7z+5+5)$$

Add $5$ and $5$ to get $10$.

$$\frac{\mathrm{d}}{\mathrm{d}z}(7z+10)$$

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 $nax^{n-1}$.

$$7z^{1-1}$$

Subtract $1$ from $1$.

$$7z^{0}$$

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

$$7\times 1$$

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

$$7$$