Question

1/5x + 7=21 - 1/20x и 8(6+x) - 4x = 5x-59

Answer 1

Ответ:

1/5x+7=21-1/20х                       8(6+х)-4х=5х- 59                                  

4х+140=420-х                          48+8х-4х=5х-59

4х+х=420-140                          48+4х=5х-59

5х=280                                     4х-5х=-59-48

х=280:5                                   -1х= -107

х=56                                          х= -107: -1

                                                  х=107

Answer 2

Ответ:

Сначала решим 1/5x + 7=21 - 1/20x

1/5x + 1/20x = 21-7

(дроби приводим к общему знаменателю - 20)

4/20x + 1/20х = 14

5/20х = 14

(Сокращаем дробь)

1/4х = 14

х = 14:1/4

х = 56

Теперь решим 8(6+x) - 4x = 5x-59

48 + 8х - 4х = 5х - 59

8х - 4х - 5х = -59 - 48

-х = -107

х = 107

Значит, 1/5x + 7=21 - 1/20x  <  8(6+x) - 4x = 5x-59

Пошаговое объяснение:

Answer 3

To solve the system of equations, let's start by solving each equation individually and then find the values of 'x' that satisfy both equations.

Equation 1: $\frac{1}{5}x + 7 = 21 - \frac{1}{20}x$

Multiply both sides of the equation by 20 to eliminate fractions: $20 \cdot \left( \frac{1}{5}x + 7 \right) = 20 \cdot (21 - \frac{1}{20}x)$

Simplify: $4x + 140 = 420 - x$

Add 'x' to both sides: $5x + 140 = 420$

Subtract 140 from both sides: $5x = 280$

Divide both sides by 5: $x = 56$

Equation 2: $8(6+x) - 4x = 5x - 59$

Distribute 8 on the left side: $48 + 8x - 4x = 5x - 59$

Combine like terms: $48 + 4x = 5x - 59$

Subtract 4x from both sides: $48 = x - 59$

Add 59 to both sides: $x = 107$

Now we have two potential values for 'x': $x = 56$ and $x = 107$. Let's check if they satisfy both equations:

For $x = 56$: Equation 1: $\frac{1}{5} \cdot 56 + 7 = 21 - \frac{1}{20} \cdot 56$ $11.2 + 7 = 21 - 2.8$ $18.2 = 18.2$ (True)

Equation 2: $8(6+56) - 4 \cdot 56 = 5 \cdot 56 - 59$ $8 \cdot 62 - 224 = 280 - 59$ $496 - 224 = 221$ (False)

For $x = 107$: Equation 1: $\frac{1}{5} \cdot 107 + 7 = 21 - \frac{1}{20} \cdot 107$ $21.4 + 7 = 21 - 5.35$ $28.4 = 15.65$ (False)

Equation 2: $8(6+107) - 4 \cdot 107 = 5 \cdot 107 - 59$ $8 \cdot 113 - 428 = 535 - 59$ $904 - 428 = 476$ (True)

It appears that only $x = 107$ satisfies both equations. Therefore, the solution to the system of equations is $x = 107$.