How to solve 10x/5-7x-5x+15-8

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 10x/5-7x-5x+15-8 | Equatix

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Question

$$10x/5-7x-5x+15-8$$

Evaluate

$7-10x$

Solution Steps

Divide $10x$ by $5$ to get $2x$.

$$2x-7x-5x+15-8$$

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

$$-5x-5x+15-8$$

Combine $-5x$ and $-5x$ to get $-10x$.

$$-10x+15-8$$

Subtract $8$ from $15$ to get $7$.

$$-10x+7$$

Show Solution

Differentiate w.r.t. x

$-10$

Steps Using Derivative Rule for Sum

Divide $10x$ by $5$ to get $2x$.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x-7x-5x+15-8)$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(-5x-5x+15-8)$$

Combine $-5x$ and $-5x$ to get $-10x$.

$$\frac{\mathrm{d}}{\mathrm{d}x}(-10x+15-8)$$

Subtract $8$ from $15$ to get $7$.

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

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 $nax^{n-1}$.

$$-10x^{1-1}$$

Subtract $1$ from $1$.

$$-10x^{0}$$

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

$$-10$$