1 )3x + 8y = 1

7x + 4y = 17
Answer: (3,1)

2) -3x + 5y = -7
9x + 2y = 38
Answer: (4,1)

3) 7r + 3s = 13
14r - 15s = -16
Answer: (1,2)

4) 2x - 7y = 42
4x - 14y = 64
Answer: No Solution ??

5) 9x + 12y = 15
3x + 4y = 5
Answer: Identity ??

6) 2 = 4s - 3t
2 = 12s - 9t
Answer: No Solution ??

7) 3a - 6b = 18
4a + 8b = 24
Answer: (6,0)

If I got any of these wrong, can someone please show me the right way to do them?

These I have no idea on how to do them.

8) 4x +8y = 5
5x +10y = 13

9) 13 = 8a - 8b
6 = 4a - 5b

10) 160 = 2x - 10y
120 = 3x - 15y

Can anyone show me how to do these ones, I keep trying to solve them, but keep getting the wrong answer?

math help needed.

There may not be tutors available for your subject and question at all times.

Have patience.
You posted this again about 1/2 hour after your first post of this question

You have to keep checking back to see if your received an answer, sometimes it might take over an hour or even more.

I gave you an answer in

http://www.jiskha.com/display.cgi?id=1235612870

I did not answer your first part, because all you have to do is check if your answers work in both of the two equations.

Sure! Let's go through each of the equations step-by-step to find the solutions:

8) To solve the system of equations:
4x + 8y = 5
5x + 10y = 13

We can simplify the equations by dividing both sides of the first equation by 4 and the second equation by 5 to make the coefficients of x the same:
x + 2y = 5/4
x + 2y = 13/5

We can see that both simplified equations are the same, meaning they represent the same line. Therefore, this system of equations has infinitely many solutions and the equations are dependent.

9) To solve the system of equations:
13 = 8a - 8b
6 = 4a - 5b

We can simplify the equations by dividing both sides of the first equation by 8 and the second equation by 4:
(8/8)a - (8/8)b = 13/8
(4/4)a - (5/4)b = 6/4

Simplifying further:
a - b = 13/8
a - (5/4)b = 3/2

Now, subtract the second equation from the first equation to eliminate a:
(a - b) - (a - (5/4)b) = (13/8) - (3/2)
Simplifying further:
(5/4)b - b = (13/8) - (3/2)

Combining like terms, we get:
-(3/4)b = (13/8) - (3/2)

Simplifying the right-hand side:
-(3/4)b = (13/8) - (12/8)
-(3/4)b = (1/8)

Dividing both sides of the equation by -(3/4):
b = -(1/8) / (-(3/4))
Simplifying further:
b = 1/6

Substitute the value of b back into one of the original equations to solve for a. Let's use the first equation:
13 = 8a - 8(1/6)
Simplifying further:
13 = 8a - 4/3
Multiply each term by 3 to eliminate the fraction:
39 = 24a - 4
Add 4 to both sides:
43 = 24a
Divide both sides by 24 to solve for a:
a = 43/24

Therefore, the solution to this system of equations is (a, b) = (43/24, 1/6).

10) To solve the system of equations:
160 = 2x - 10y
120 = 3x - 15y

We can simplify the equations by dividing both sides of the first equation by 2 and the second equation by 3 to make the coefficients of x the same:
x - 5y = 80
x - 5y = 40

We can see that both simplified equations are the same. Therefore, this system of equations has infinitely many solutions and the equations are dependent.

I hope this helps! Let me know if you have any further questions.

Sure! Let me go through each problem and explain how to solve them.

1) To solve this system of equations:
3x + 8y = 1 [Equation 1]
7x + 4y = 17 [Equation 2]

One way to solve it is by using the method of substitution. We can solve Equation 1 for x in terms of y:
3x = 1 - 8y
x = (1 - 8y) / 3

Then substitute this expression for x into Equation 2:
7((1 - 8y) / 3) + 4y = 17

Now, you can solve this equation algebraically to find the value of y. Substitute this value back into either Equation 1 or Equation 2 to solve for x. The solution is (3, 1).

2) Following the same steps as before:
-3x + 5y = -7 [Equation 1]
9x + 2y = 38 [Equation 2]

Solve Equation 1 for x:
-3x = -7 - 5y
x = (-7 - 5y) / -3

Substitute this expression for x into Equation 2:
9((-7 - 5y) / -3) + 2y = 38

Solve this equation to find the value of y. Then substitute back into either original equation to solve for x. The solution is (4, 1).

3) Again, using the substitution method:
7r + 3s = 13 [Equation 1]
14r - 15s = -16 [Equation 2]

Solve Equation 1 for r:
7r = 13 - 3s
r = (13 - 3s) / 7

Substitute this expression for r into Equation 2:
14((13 - 3s) / 7) - 15s = -16

Solve for s and then substitute back into either Equation 1 or Equation 2 to solve for r. The solution is (1, 2).

4) This system of equations:
2x - 7y = 42 [Equation 1]
4x - 14y = 64 [Equation 2]

We can see that Equation 2 is simply twice Equation 1. These two equations represent the same line, so they are parallel and will not intersect. Therefore, there is no solution.

5) The equations:
9x + 12y = 15 [Equation 1]
3x + 4y = 5 [Equation 2]

If you multiply Equation 2 by 3, you get:
9x + 12y = 15

This means that both equations represent the same line, and every point on that line will be a solution. In other words, the system of equations has infinitely many solutions, which is called an identity.

6) The equations:
2 = 4s - 3t [Equation 1]
2 = 12s - 9t [Equation 2]

If you divide Equation 2 by 2, you get:
1 = 6s - 4.5t

This equation can't be true because the coefficients of s and t are different, but the constants are the same. Therefore, the system of equations has no solution.

7) For the equations:
3a - 6b = 18 [Equation 1]
4a + 8b = 24 [Equation 2]

By dividing Equation 1 by 3 and Equation 2 by 4, we get:
a - 2b = 6
a + 2b = 6

By eliminating the variable b, we add these two equations:
2a = 12
a = 6

Substitute the value of a into either equation to solve for b. The solution is (6, 0).

Now moving on to the unanswered problems:

8) For the equations:
4x + 8y = 5 [Equation 1]
5x + 10y = 13 [Equation 2]

You can see that Equation 2 is simply Equation 1 multiplied by 5/4. These two equations represent the same line, so they have infinitely many solutions, which is an identity.

9) The equations:
13 = 8a - 8b [Equation 1]
6 = 4a - 5b [Equation 2]

To solve this system, you can multiply Equation 1 by 5 and Equation 2 by 8 to eliminate b:
65 = 40a - 40b
48 = 32a - 40b

Subtracting Equation 2 from Equation 1, we get:
65 - 48 = 40a - 32a - 40b + 40b
17 = 8a

Solve for a and substitute this value back into either original equation to solve for b.

10) The equations:
160 = 2x - 10y [Equation 1]
120 = 3x - 15y [Equation 2]

First, divide Equation 1 by 2 and Equation 2 by 3 to eliminate x:
80 = x - 5y
40 = x - 5y

Now we have two equations that represent the same line. This means they have infinitely many solutions, which is an identity.

I hope this helps you understand how to approach each problem!