Math problem help please?

Question

For each system, state multipliers you would use to eliminate each of the variables by addition.

x - y = 1
3x - y = 3


Solve each system of equations by the substitution method

x - 2y = 5
3x - 5y = 8

Answer 1

x - y = 1- - - - - - -Equation 1
3x - y = 3- - - - - -Equation 2
- - - - - - -

Multiply equation 1 by - 1

-1(x) - (-1)(y) = -1(1)

-x + y = - 1

- - - - - - - - - - - -

-x + y = - 1
3x - y = 3
- - - - - - - -

2x = 2

2x/2 = 2/2

x = 1

The answer is x = 1

Insert the x value into equation 2

- - - - - - - - - - - - - - - - - - - - - -

3x - y = 3

3(1) - y = 3

3 - y = 3

2 - y - 3 = 3 - 3

- y = 0

-1(- y) = -1 (0)

y = 0

The answer is y = 0

Insert the y value into equation 2

- - - - - - - - - - - - - - - - - - - - - -

3x - y = 3

3(1) - 0 = 3

3 = 3

- - - - - - - - - -

x - y = 1

1 - 0 = 1

1 = 1

- - - - - - - - -

The solution set is { 1, 0 }

- - - - - - - -s-

Answer 2

For each system, state multipliers you would use to eliminate each of the variables by addition.

x - y = 1
3x - y = 3

To eliminate the y's you could multiply either equation through by -1 (note, I am assuming strictly addition. In this instance one would usually just subtract the first equation from the second)
as -(-y) = y and -y + y = 0

It would either leave you with
-2x = -2 or 2x = 2, both of which leave x = 1

To eliminate x's you can multiply the first equation through by -3 (again, this is only for strictly addition. Subtracting would be the more sensible function here)
-3x + 3y = -3
3x - y = 3

to give 2y = 0 or y = 0

Your answer is x = 1, y = 0
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Solve each system of equations by the substitution method

x - 2y = 5
3x - 5y = 8

Rearranging the first equation
x = 2y + 5
Substituting into the second equation
3(2y + 5) - 5y = 8
Expand all brackets
6y + 15 - 5y = 8
Put like terms together
y + 15 = 8
y = -7

(Note: at this point, the most sensible option would be to say x - 2(-7) = 5 so x = -9)

x - 2y = 5
so 2y = x - 5
so y = (x - 5)/2

3x - 5y = 8
3x - 5((x - 5)/2) = 8
6x - 5(x - 5) = 16
6x - 5x + 25 = 16 (note: -5 * -5 = + 25)
x + 25 = 16
x = -9

Your answer is x = -9, y = -7

Answer 3

Solution I:
x -y = 1 multiply by 3 the coefficient of 3x in the second equation.
we get, first equation as 3x - 3y = 3
and second equation is 3x - y = 3
By subtracting second equation from first equation,
we get, -2y = 0
Or y = 0

Now if, y = 0, then x = 1 by substituting the value of y in first equation.
Hence the solution set is ( x = 1, y = 0 ).

Solution II:
x = 5 + 2y substitute this value in second equation, we get
15 + 6y -5y = 8
Or y = 8 - 15 = -7 Therefore x = 5 - 14 = -9
The solution set is (x = -9, y = -7)

Answer 4

1. Multiply the first equation by -3
-3x + 3y = -3
3x - y = 3
Add them together and you will get: 2y = 0; y = 0

2. Multiply the second equation by -1
x - y = 1
-3x + y = -3
Add them together and you will get -2x = -2; x = 1

3. Substitution 1
x = y+1; 3(y+1)=3; y=0; x=1

4. Substituation 2
x = 2y+5; 3(2y+5)-5y=8; y+15=8; y=-7; x=-9

Answer 5

multiply x - y = 1 by "-3" to eliminate x by adding the two


or to eliminate y multiply throughout by " -1"







for second part


x - 2y = 5
so x=5+2y


substituting we get


3*(5+2y) - 5y = 8

15 + 6y - 5y = 8
y = 8 - 15

y = -7


substituting again

x - 2 * ( -7 ) = 5
x+14=5
x=5-14
x=-9

Answer 6

x - y = 1; multiply by -3
3x - y = 3;
add them : 3*y - y= 0; y=0
x - y = 1; multiply by -1
3x - y = 3;
add them : 2x =2 ; x=1;

x-2*y=5;
x = 5+2*y; 3(5+2*y) - 5*y =8;
15+y = 8; y = -7;
replace again x - 2*-7=5; x = 5-14 =-9;

Answer 7

for the x-2y=5 the x would be 7 minus 2 which would be 5 and than times it by one and you get five.
i have to go, but best of luck on the other ones.

Answer 8

the 1st
x=1+y

3(1+y)-y=3
3+2y=3
y=0

x=1


the 2nd(it has no solution)


x=5+2y

3(5+2y)-5y=8
15-y=8
y=7

x-14=5
x=19 but 3*19-5*7=8 ????!!!!

Answer 9

it is tooo hard