How do you solve this system of equations? 2x+y=1 and x-y=8?

Answer 1

Add the two equations together: (2x+y)+(x-y) = 1+8, so 3x = 9, x = 3. Then 2*3+y = 1 so y = -5.

Steve

Answer 2

there's a couple different ways, the way I find easiest is to solve for one variable in terms of the other:
2x+y=1 and x-y=8 (I'll use the second one and x)
x-y=8 -- x=y+8
now put that into the other equation
2(y+8)+y=1
then solve for y
2y+16+y=1 -- 3y=-15 -- y=-5
then put that into an equation to solve for x (doesn't matter which equation)
x-(-5)=8 -- x=3
that gives you y=-5 and x=3 you can just insert them into the other equation if you have to check your work.

Answer 3

by substitution: replace y by x - 8 in the first equation and then you will find 3x = 9 then x = 3 y = - 5

or by summing the two equations

there is a general algorithm called teh Gauss pivot to solve that kinf of system provided certain conditions

Answer 4

2x+y=1
x-y=8
3x=9
x=3

These are called simultaneous equations the third line above is the sum of the first two equations eliminating y and getting a value for x

To find y sub x in to any of the equations

3-y=8
y=-5

Answer 5

2x+y=1
x-y=8

x=8+y
subsitute x to get
2(8+y)+y=1

16+2y+y=1
16+3y=1
3y=1-16
3y=-15
y= -5


then x should be

x-(-5)=8
x+5=8
x=8-5
x=3

Answer 6

2x+y=1
x-y=8

x=8+y

2(8+y)+y=1
16+2y+y=1
3y=-15
y=-5

x=8+(-5)
x=3

Answer 7

x-y=8
x=8+y

2x+y=1
2(8+y)+y=1
16+2y+y=1
16+3y=1
3y=1-16
3y=-15
y=-5

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