Answer :
an equvilent equation
remember you can do anything to an equation as long asyou do it to both sides
assuming yo have
x+y=1 and
x-3y=9
mulitply both by 2
2x+2y=2
2x-6y=18
those are equvilent
ok, solve initial
x+y=1
x-3y=9
multiply first equation by -1 and add to 2nd equation
-x-y=-1
x-3y=9 +
0x-4y=8
-4y=8
divide both sides by -4
y=-2
sub back
x+y=1
x-2=1
add 2
x=3
x=3
y=-2
(3,-2)
if we test it in other one
2x+2y=2
2(3)+2(-2)=2
6-4=2
2=2
yep
2x-6y=18
2(3)-6(-2)=18
6+12=18
18=18
yep
solution is (3,-2)
remember you can do anything to an equation as long asyou do it to both sides
assuming yo have
x+y=1 and
x-3y=9
mulitply both by 2
2x+2y=2
2x-6y=18
those are equvilent
ok, solve initial
x+y=1
x-3y=9
multiply first equation by -1 and add to 2nd equation
-x-y=-1
x-3y=9 +
0x-4y=8
-4y=8
divide both sides by -4
y=-2
sub back
x+y=1
x-2=1
add 2
x=3
x=3
y=-2
(3,-2)
if we test it in other one
2x+2y=2
2(3)+2(-2)=2
6-4=2
2=2
yep
2x-6y=18
2(3)-6(-2)=18
6+12=18
18=18
yep
solution is (3,-2)
equivalent system is the same set of eqns n has the same solution
so for x + y = 1 and x + 3y = 9
an equivalent system can be created by multiplying everything by 3
3x + 3y = 3 and 3x + 9y = 27
solving by substracting one from the other
9y - 3y = 27 - 3
6y = 24
y = 4
x = 1 - y = 1 - 4
= -3
substituting (-3,4) back into
x + y = 1 and x + 3y =9
-3 + 4 = 1 and -3 + 3*4 = 9
so the solutions are good
so for x + y = 1 and x + 3y = 9
an equivalent system can be created by multiplying everything by 3
3x + 3y = 3 and 3x + 9y = 27
solving by substracting one from the other
9y - 3y = 27 - 3
6y = 24
y = 4
x = 1 - y = 1 - 4
= -3
substituting (-3,4) back into
x + y = 1 and x + 3y =9
-3 + 4 = 1 and -3 + 3*4 = 9
so the solutions are good