How to solve y-2-3-4-5-6-7

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 y-2-3-4-5-6-7 | Equatix

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Question

$$y-2-3-4-5-6-7$$

Evaluate

$y-27$

Solution Steps

Subtract $3$ from $-2$ to get $-5$.

$$y-5-4-5-6-7$$

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

$$y-9-5-6-7$$

Subtract $5$ from $-9$ to get $-14$.

$$y-14-6-7$$

Subtract $6$ from $-14$ to get $-20$.

$$y-20-7$$

Subtract $7$ from $-20$ to get $-27$.

$$y-27$$

Show Solution

Differentiate w.r.t. y

$1$

Steps Using Derivative Rule for Sum

Subtract $3$ from $-2$ to get $-5$.

$$\frac{\mathrm{d}}{\mathrm{d}y}(y-5-4-5-6-7)$$

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

$$\frac{\mathrm{d}}{\mathrm{d}y}(y-9-5-6-7)$$

Subtract $5$ from $-9$ to get $-14$.

$$\frac{\mathrm{d}}{\mathrm{d}y}(y-14-6-7)$$

Subtract $6$ from $-14$ to get $-20$.

$$\frac{\mathrm{d}}{\mathrm{d}y}(y-20-7)$$

Subtract $7$ from $-20$ to get $-27$.

$$\frac{\mathrm{d}}{\mathrm{d}y}(y-27)$$

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 $nax^{n-1}$.

$$y^{1-1}$$

Subtract $1$ from $1$.

$$y^{0}$$

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

$$1$$