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. Solve the system of equations using the addition (elimination) method.
If the answer is a unique solution, present it as an ordered pair: (x, y). If not, specify whether the answer is “no solution”
or “infinitely many solutions.”
x + y = 4
2x – y = 2



Solve the system of equations using the addition (elimination) method.
If the answer is a unique solution, present it as an ordered pair: (x, y). If not, specify whether the answer is “no solution” or “infinitely many solutions.”
5x – 3y = 3
-10x + 6y = -4

2007-04-12 06:27:15 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

4 answers

x+y=4
2x-y=2
adding them both you get

3x=6
x=2
y=4-2
y=2

(2,2)

the second one is not possible

simplify the second equation

-1/2{(-10x+6y)=-4}

5x-3y=2
one equation can not have 2 different solutions.

2007-04-12 06:33:19 · answer #1 · answered by Anonymous · 0 0

1. x + y = 4
2x – y = 2

add the 2 equations:
3x = 6
x = 2

substitute in x to the 1st equation:
2 + y = 4
y = 2

(2,2) is your answer.

2. 5x – 3y = 3
-10x + 6y = -4

multiply the 1st equation by 2:
10x - 6y = 6

then add the 2 equations together:
0x + 0y = -1

i think it's no solution.

2007-04-12 06:35:37 · answer #2 · answered by Pluie 2 · 0 0

x + y = 4- - - - - -Equation 1
2x - y = 2- - - - - Equation 2
- - - - - - - -
3x = 6

3x / 3 = 6/3

x = 6/3

x = 2

Insert the x value into euation 1
- - - - - - - - - - - - - - - - - -- - - - - -

x + y = 4

2 + y = 4

2 + y - 1 = 4 - 2

y = 2

Insert the y value into equation 1

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

Check for equation 1

x + y = 4

2 + 2 = 4

4 = 4

- - - - - -

Check for equation 2

2x - y = 2

2(2) - 2 = 2

4 - 2 = 2

2 = 2

- - - - - - - - - - -

Both equations balance

The solution set is { 2, 2 }

- - - - - - - - - - s-

2007-04-12 07:12:01 · answer #3 · answered by SAMUEL D 7 · 0 0

1. Add them together:
3x = 6 ------> x = 2 ; ------> y =2
(2,2)

2. divided (2) by 2 and add them together
0x + 0y = 1
No solution

2007-04-12 06:35:24 · answer #4 · answered by tuoidabuon 2 · 0 0

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