Question

1. What is the solution of the system?

− + 2 = 4
−4 + = −5

2. How many solutions does the system have?
−2 + = 3
4 − 4 =

3. How many solutions does the system have?
3 + = 5
−2 = 6 − 10

4. How many solutions does the system have?
− − 2 = −14
4 = −2 − 12

6. Solve the following system using substitution or elimination.
3 + 3 = 27
− 3 = −11

7. Solve the following system using substitution or elimination.
2 + 3 = 9
+ 5 = 8

and if you don't mind could you explain how you got your answers

Answer 1

The given system of equations has solutions below:

1) The solution is (2,3)

2) The solution is (
`(-8)/(7), (5)/(7)`)

3) The solution is infinitely many solutions

4) No solution

Explanation:

Given system of equation are

`-x+2y=4\hfill (1)`

`-4x+y=-5\hfill (2)`

To solve equation by using elimination method

Multiply eqn (2) into 2

`-8x+2y=-10\hfill (3)`

Now subtracting (1) and (3)

`-x+2y=4`

`-8x+2y=-10`

_________________

7x=14

x=
`(14)/(7)`

`x=2`

Substitute x=2 in equation (1)

-2+2y=4

2y=4+2

`y=(6)/(2)`

y=3

Therefore the solution is (2,3)

2) Given equation is

`-2x+y=3\hfill (1)`

4y-4=x

Rewritting as below

`x-4y=-4\hfill (2)`

To solve equation by using elimination method

multiply (2) into 2

`2x-8y=-8\hfill (3)`

Adding (1) and (3)

-2x+y=3

2x-8y=-8

________

-7y=-5

`y=(5)/(7)`

substitute
`y=(5)/(7)` in (1)

`-2x+(5)/(7)=3`

`-2x=3-(5)/(7)`

`-2x=(21-5)/(7)`

`x=-(8)/(7)`

Therefore the solution is (
`(-8)/(7),(5)/(7)`)

3) Given equation is
`6x+2y=10\hfill (1)`

`3x+y=5\hfill (2)`

equation (1) can be written as

2(3x+y)=10

`3x+y=(10)/(2)`

3x+y=5

Therefore equations (1) and (2) are same therefore it has infinitely many solutions

4) Given equation is
`-x-2y=14\hfill (1)`

`-2x-4y=12\hfill (2)`

multiply equation (1) into 2

`-2x-4y=28\hfill (3)`

To solve equation by using elimination method

subtracting (2) and (3)

-2x-4y=28

-2x-4y=12

_______

`28\\eq -12`

therefore it has no solution