Question

1257. 1)8x + 7 y + 53 = 0,4x – 5y + 1 = 0;2)7 x + 9y + 48 = 0,6x + 3y + 27 = 0.​

Answer 1

1)8х+7у+53=0
4х-5у+1=0

8х+7у=-53
4х-5у=-1. *(-2)

8х+7у=-53
-8х+10у=2

17у=-51
-8х+10у=2

У=-3
Х=-4
2)7х+9у+48=0
6х+3у+27=0

7х+9у=-48
6х+3у=-27. (*-3)

7х+9у=-48
-18х-9у=81

-11х=33
7х+9у=-48

х=-3
у=-3

Answer 2

It appears that you have provided two sets of linear equations, and it seems like you want to solve these systems of equations. Let's solve each system one by one:

System 1:

1. 8x + 7y + 53 = 0
2. 4x - 5y + 1 = 0

We can solve this system using either the substitution method or the elimination method. I'll use the elimination method.

First, let's multiply the second equation by 2 to make it easier to eliminate one of the variables:

2 * (4x - 5y + 1) = 2 * 0 8x - 10y + 2 = 0

Now, let's add the two equations together to eliminate the x variable:

(8x + 7y + 53) + (8x - 10y + 2) = 0

This simplifies to:

16x - 3y + 55 = 0

Now, let's isolate the y variable in this new equation:

16x - 3y + 55 = 0

-3y = -16x - 55

Divide both sides by -3:

y = (16/3)x + 55/3

Now, you have the value of y in terms of x. You can substitute this expression for y into one of the original equations to find the value of x, or you can keep it in this form.

System 2:

1. 7x + 9y + 48 = 0
2. 6x + 3y + 27 = 0

Again, we can use the elimination method. First, let's multiply the second equation by 3 to make it easier to eliminate one of the variables:

3 * (6x + 3y + 27) = 3 * 0 18x + 9y + 81 = 0

Now, let's subtract the first equation from the new equation to eliminate the y variable:

(18x + 9y + 81) - (7x + 9y + 48) = 0

This simplifies to:

11x + 33 = 0

Now, isolate the x variable:

11x = -33

x = -33/11

x = -3

Now that you have the value of x, you can substitute it into one of the original equations to find the value of y:

7x + 9y + 48 = 0 7(-3) + 9y + 48 = 0 -21 + 9y + 48 = 0 9y + 27 = 0

Now, isolate the y variable:

9y = -27

y = -27/9

y = -3

So, the solution to the second system is x = -3 and y = -3.