Question

6(x-3)=7y-1 2(y+6)=32+2​

Answer 1

Объяснение:

2.

2(у+6)=32+2

2у+12=34

2у= 34-12

2у= 22

у=22-2=11

ответ: 11

1.

6(х-3)=7у-1

6х-18=7у-1

6х=7у-1+18

6х=7у+17

х=7у/6+ 17/6

ответ: 7у/6 +17/6

Answer 2

To solve this system of equations, you can follow these steps:

Equation 1: 6(x - 3) = 7y - 1
Equation 2: 2(y + 6) = 32 + 2

Let's solve for y in each equation and then find the corresponding x value.

Equation 1: Expand the left side of the equation: 6x - 18 = 7y - 1

Now, isolate y by dividing both sides of the equation by 7: 7y = 6x - 18 + 1
7y = 6x - 17
y = (6x - 17) / 7

Equation 2: Expand the left side of the equation: 2y + 12 = 32 + 2

Now, isolate y by subtracting 12 from both sides of the equation: 2y = 32 + 2 - 12
2y = 22
y = 22 / 2
y = 11

Now that you have the value of y from both equations, you can substitute it back into either Equation 1 or Equation 2 to solve for x.

Using Equation 1: y = (6x - 17) / 7
11 = (6x - 17) / 7

Multiply both sides by 7 to get rid of the fraction: 77 = 6x - 17

Add 17 to both sides: 6x = 94

Divide both sides by 6: x = 94 / 6
x = 47 / 3

So, the solution to the system of equations is: x = 47 / 3
y = 11