Question

5-7*(4+x)=10-4x (7-2x)*(5x-1)=0

Answer 1

Ответ:

5-7*(4+х)=10-4х

5-28-7х=10-4х

-7х+4х=10-5+28

-3х=33

х=33÷3

х=11

(7-2х)*(5х-1)=0

35х+7-10х+2х=0

35х-10х+2х=0-7

27х=-7

х=-7÷27

х=-7/27

Answer 2

To solve the given system of equations, let's start by solving each equation separately.

Equation 1: $5 - 7 \cdot (4 + x) = 10 - 4x$

First, distribute the -7 on the left side:

$5 - 28 - 7x = 10 - 4x$

Combine like terms:

$-23 - 7x = 10 - 4x$

Move the terms involving $x$ to one side and constants to the other side:

$-23 + 10 = 4x - 7x$

$-13 = -3x$

Divide by -3:

$x = \frac{13}{3}$

Equation 2: $(7 - 2x) \cdot (5x - 1) = 0$

Using the zero-product property, either $7 - 2x = 0$ or $5x - 1 = 0$.

For the first case:

$7 - 2x = 0$

$-2x = -7$

$x = \frac{7}{2}$

For the second case:

$5x - 1 = 0$

$5x = 1$

$x = \frac{1}{5}$

Therefore, the solutions to the system of equations are $x = \frac{13}{3}$, $x = \frac{7}{2}$, and $x = \frac{1}{5}$.