Question

18-4y=7(2-y)+6; 4(-2z+5)=14-2(4z-3); 5(n-8)-3(4-2n)=7(3n-7)+9; -6x+2(5-3x)=8;

Answer 1

Пошаговое объяснение:

1)

18-4у=7(2-у)+6

18-4у=14-7у+6

-4у+7у=14+6-18

3у=2

$y = \frac{2}{3}$

2)

4(-2z+5)=14-2(4z-3)

-8z+20=14-8z+6

-8z+8z=14+6-20

0=0

z Є R

3)

5(n-8)-3(4-2n)=7(3n-7)+9

5n-40-12+6n=21n-49+9

21n-5n-6n=49-9+12+40

10n=92

$n = \frac{92}{10} = 9.2$

4)

-6x+2(5-3x)=8

-6x+10-6x=8

-6x-6x=8-10

-12x=-2

12x=2

$x = \frac{2}{12} = \frac{1}{6}$

Answer 2

Let's solve each of these equations step by step:

1. $18 - 4y = 7(2 - y) + 6$:

First, distribute the 7 on the right side:

$18 - 4y = 14 - 7y + 6$

Combine like terms:

$18 - 4y = 20 - 7y$

Now, add 7y to both sides to isolate y:

$18 + 3y = 20$

Subtract 18 from both sides:

$3y = 20 - 18$

$3y = 2$

Finally, divide both sides by 3 to solve for y:

$y = \frac{2}{3}$

So, the solution is $y = \frac{2}{3}$.

2. $4(-2z + 5) = 14 - 2(4z - 3)$:

First, distribute the 4 and -2 on the left and right sides, respectively:

$-8z + 20 = 14 - 8z + 6$

Now, notice that the terms with -8z on both sides will cancel out when we subtract them:

$-8z + 20 = -8z + 20$

This equation simplifies to:

$20 = 20$

The equation is an identity, meaning it's true for all values of z. So, there is no unique solution for z.

3. $5(n - 8) - 3(4 - 2n) = 7(3n - 7) + 9$:

First, distribute on both sides:

$5n - 40 - 12 + 6n = 21n - 49 + 9$

Combine like terms:

$11n - 52 = 21n - 40$

Now, subtract 11n from both sides:

$-52 = 10n - 40$

Add 40 to both sides:

$-52 + 40 = 10n$

$-12 = 10n$

Finally, divide both sides by 10 to solve for n:

$n = -\frac{12}{10}$

Simplify the fraction:

$n = -\frac{6}{5}$

So, the solution is $n = -\frac{6}{5}$.

4. $-6x + 2(5 - 3x) = 8$:

First, distribute the 2 on the left side:

$-6x + 10 - 6x = 8$

Combine like terms:

$-12x + 10 = 8$

Subtract 10 from both sides:

$-12x = 8 - 10$

$-12x = -2$

Finally, divide both sides by -12 to solve for x:

$x = \frac{-2}{-12}$

Simplify the fraction:

$x = \frac{1}{6}$

So, the solution is $x = \frac{1}{6}$.