math question ?

Question

how do u do this.


1. x-y=9
x+y=7

2. 5x-4y=12
3x-4y=4

3. 3x+3y=6
2x-y=1

4. 3x-5y=13
4x+3y=17

5. x-y=1
3x-y=10

6. 4x=2y-7

7. y=-2x
5x+3y=-3

8. x+y=6
-2x+y=-3

9. 2x+y=45
3x-y=5

i don't get this stuff at all please help me

Answer 1

1. x-y=9---------------------(1)
x+y=7-----------------------(2)
adding
2x=16
dividing by 2
x=8
sub in 1,
y=-1

2. 5x-4y=12--------(1)
3x-4y=4-------------(2)
(1)-(2)
2x=8
dividing by 2
x=4
sub in 1
y=2

3. 3x+3y=6-----------(1)
2x-y=1-----------------(2)*3
6x-3y=3---------------(3)
(1)+(3)
9x=9
dividing by 9
x=1
sub in (1)
y=1

4. 3x-5y=13-----------(1)*3
4x+3y=17--------------(2)*5
9x-15y=39
20x+15y=85
adding
29x=124
dividing by 29
x=124/29
sub and find y

5. x-y=1---------(1)
3x-y=10---------(2)
(2)-(1)
2x=9
dividing by 2
x=9/2
suby=7/2

6. 4x=2y-7
the other equation is missing
x=(2y-7)/4

7. y=-2x-----------(1)
5x+3y=-3---------(2)
sub y=-2x in (2)
5x+3(-2x)=-3
5x-6x=-3
-x=-3
x=3
sub
y=-6

8. x+y=6---------(1)
-2x+y=-3---------(2)
subtracting
3x=9
dividing by 3
x=3
sub
y=3

9. 2x+y=45-----------(1)
3x-y=5-----------------(2)
adding
5x=50
dividing by 5
x=10
sub
y=25

Answer 2

1. add the equations together... y cancels out so your left with the answer for x
x - y = 9
x + y = 7
2x + 0y = 16
x = 8
plug this number back in one of the equations to find y
8 - y = 9
y = -1
2. 5x-4y=12 5x-4y=12
(3x-4y=4)(-1) -3x + 4y = -4
2x = 8
x = 4, y = 2
3. 3x+3y=6
2x-y=1 set this equation equal to y
y = 2x - 1 plug this equation in for y in the first equation
3x + 3(2x - 1) = 6
3x + 6x - 3 = 6
9x = 9
x = 1, y = 1

4. (3x-5y=13)3 9x - 15y = 39
(4x+3y=17)5 20x + 15y = 85
29x = 124
x = 124/29, y = -1/29
5. x-y=1 x = y + 1
3x-y=10
3(y + 1) - y = 10
3y + 3 - y = 10
2y = 7
y = 7/2, x = 9/2

6. not enough information

7. y=-2x
5x+3y=-3
5x + 3(-2x) = -3
5x - 6x = -3
x = 3, y = -6

8. (x+y=6)2 2x + 2y = 12
-2x+y=-3 -2x+y=-3
3y = 9
y = 3, x = 3

9. 2x + y = 45
3x - y = 5
5x = 50
x = 10, y = 25

Answer 3

I assume you want to solve the systems of equations.

1.
1a) x - y = 9
1b) x + y = 7

If you add the equations together, you get

2x = 16
x = 8 {this result by the method of addition}

Now put this value of x into either original equation to find the corresponding value for y

1) x + y = 7
8 + y = 7
y = -1

Another approach is to solve one of the equations for one of the variables

1a) x - y = 9
x = y + 9

Then replace x with y - 9 in the other equation

1a) x + y = 7
y + 9 + y = 7
2y + 9 = 7
2y = -2
y = -1 {this result by the method of substitution}


2.
2a) 5x - 4y = 12
2b) 3x - 4y = 4

This calls for the method of subtraction

5x - 3x - 4y - (-4y) = 12 - 4
2x = 8
x = 4

2a) 5x - 4y = 12
5(4) - 4y = 12
20 - 4y = 12
-4y = -8
y = 1/2


3.
3a) 3x + 3y = 6
3b) 2x - y = 1

Here you can use substitution - solve 3b for y and substitute for y in 3a

Another approach would be to multiply equation 3b by 3 and then add the equations

3x + 3y = 6
6x - 3y = 3

9x = 9
x = 1 {result by addition}

3(1) + 3y = 6
3 + 3y = 6
3y = 3
y = 1

This should get you started. Use whatever method seems appropriate, and be careful

Answer 4

1) add the two equations and get 2x = 16, so x=8.Now put x =8 into one of the two equations and solve for y. Using the second equation (x+y)=7 8+y=7 y=-1

2)Multiply the second equation by -1 getting -3x + 4y = -4
Now when you add this new equation to the first equation, the y values cancel out and you ca solve for x as shown in problem 1.

3) Multiply second equation by 3. This gets rid of y when you add.

4) Multiply top equation by 3 and bottom by 5. adding the two new equations gets rid of y letting you solve for x.

The rest of the problems are solved in a similar manner.

Answer 5

solve by substitution

1. x-y=9
x+y=7

just add the 2 problems so that you cancel out one of the variavles and you end up with

2x = 16
x = 8

now that you know what x is just plug it into either of the original problems to find the y.

(8) - y = 9
-y = 1
y = -1

the rest should be easy for you.


For 2. 5x-4y=12
3x-4y=4, the only way to cancel out is to multiple one of the problems by (-1) to cancel out the y.

-1(5x-4y=12) = -5x + 4y = -12, and now 4y and -4y cancels out

-5x + 4y = -12
3x - 4y = 4

add the 2 problems and you get

-2x = -8
x = 4

plug in the x variable to one of the problems

3(4) - 4y = 4
12 - 4y = 4
-4y = -16
y = 4

i hope this helps

Answer 6

Use algebra.

For instance, you can make equation 9 into:

-3(2x+y) = 45
2(3x - y) = 5

Multiply through to get:

-6x - 3y = 135
6x - 2y = 10

The x's cancel to give:

-5y = 145
y = -29

Now you can solve for x.

Answer 7

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Answer 8

you have to multiply so that one of the variables will cancel...
example:

8) x+y=6
-2x+y=5

you can either get rid of the x's or the y's. y is easiest.

multiply the top one by negative one.

add together
-x-y=-6
-2x+y=5
------------
-3x= -1
the y's cancel
x=1/3
put the x answer you get into either equation
the trick is to multiply so that either x or y will cancel
also if you get 0= any number bsides 0 , then the answer is no solution, but if you get to numbers equal to each other, then it is all real numbers.

Answer 9

you line up the problem and subtract. for instance
x-y=9
x+y=7
2x=16
x=8
y=-1
if the number has a coefficient higher than one, muliply
x-y=1
3x-y=10

3x-3y=3
3x-y=10
-2y=-7
y=2.5
x=3.5

Answer 10

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