Question

What is the solution of the system of equations `y=-x+2` and `y=3x-2`?

Answer 1

`(1,1)`

Explanation:

`color(red)(y)=-x+2to(1)`

`color(red)(y)=3x-2to(2)`

`"since both equations express y in terms of x we can"`
`"equate them"`

`rArr3x-2=-x+2`

`"add x to both sides"`

`3x+x-2=cancel(-x)cancel(+x)+2`

`rArr4x-2=2`

`"add 2 to both sides"`

`4xcancel(-2)cancel(+2)=2+2`

`rArr4x=4`

`"divide both sides by 4"`

`(cancel(4) x)/cancel(4)=4/4`

`rArrx=1`

`"substitute this value into either of the 2 equations"`

`x=1to(1)toy=-1+2=1rArr(1,1)`

`color(blue)"As a check"`

`x=1to(2)toy=3-2=1rArr(1,1)`

`rArr"the point of intersection " =(1,1)`
graph{(y-3x+2)(y+x-2)=0 [-10, 10, -5, 5]}

Answer 2

`x = 1, y = 1`

Explanation:

Complex linear systems can be solved in matrix form using Cramer's Rule. Simple ones like this one can be arranged according to their factors and solved algebraically.

Arrange the equations so that the factors align, with all of the unknowns on one side:
`y = −x + 2`
`y = 3x − 2`

`y + x = 2`
`y – 3x = -2`

Then algebraically combine them. You can use multiplicative factors to an entire equation if the coefficients are not already equal. Then we can simply subtract one equation from the other to get a single equation in only the 'x' variable.

`y + x = 2`
`y – 3x =-2` Subtract (1) from (2):

`-4x = -4` ; `x = 1`

Substitute this value back into one equation to solve for 'y', then use the other equation to check the final values for correctness.
`y = −1 + 2` ; `y = 1`

CHECK:
`y = 3x − 2` ; `1 = 3*1 - 2` `1= 3 - 2` ; `1 = 1` ; CORRECT!