Solving Systems Algebraically?

Question

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

2. 2x+5y=8
-3x+y=5

3. x+3y=5
2x-y=5


4. 3x-2y=6
-6x+4y=-12

5. x+5y=8
-2x-10y=4

Answer 1

1. x=4
2. x=-1
3. x=20/7
4. Equations are = so x = anything
5. DNE or Does Not Exist

Hope this helps!

Answer 2

Just add them togeyher and get 3x = 12 so x =4. Since x-y = 3, y = 1.

Multiply 2nd equation by - 5 and add to 1st equation:
15x -5y = -25 added to 2x+5y = 8 gives 17 x = -17 so x=-1.
put x = -1 in 1st equation getting 2(-1)+5y = 8 so y = 2

Multiply 2nd equation by 3 and add to 1st equation getting:
6x-3y = 15 added to x+3y = 5 gives 7x = 20 so x = 20/7.
20/7 +3y= 5, so y = (5-20/7)/3 = 15/21= 5/7

Multiply1st equation by 2 and add to 2nd equation getting:
6x-4y=12 added to -6x +4y = -12 giving 0=0. Thus they are the same equation and hence have an infinite number of solutions.

Multiply1st equation by 2 and add to 2nd equation getting:
2x+10y = 8 added to -2x -10y = 4 getting 0 = 12. Hence the lines are parallel and thus there is no solution.

Answer 3

Sorry approximately your math instructor! So what you do on the 1st one is....you're the two using removal or substitution i'm guessing with the help of the situation. for this reason you have 2 variables that are ones so substitution could be greater handy. So on the 1st (or 2d. despite you elect) positioned y on one area of the = sign and the different stuff on the different area. So ... y= -3x +sixteen. I basically moved the 3x over. then you definitely replace -3x+sixteen into the different equation for y. 2x + (-3x+sixteen) = 11. Now resolve normally. So... (2x + -3x is -x with the help of ways) -x +sixteen = 11. Subtract and you get -x= -5. exchange that to x=5. this is the same ingredient. Now all you prefer is the y. So plug the 5 into certainly one of the two equations and resolve returned. i visit apply the 2d. 2(5) + y = 11. So y= a million. good success and that i wish I helped!

Answer 4

i will show you a couple ways to do some of these:

1. This is by substitution:
2nd eqn: x = 3+y.
Plug in to the 1st eqn: 2(3+y)+y=9
6+2y+y=9
3y=3
y=1
Plug into 2nd eqn:
x-(1)=3
x=4
ANS: (4, 1)


2. This I will do by elimination:

multiply the 2nd eqn by -5 and add the two eqns:

2x+5y=8
+15x-5y=-25
----------------
17x=-17
x=-1
Plug in to 2nd eqn:
-3(-1)+y=5
y = 2
ANS: (-1, 2)

you try.......

Answer 5

1. add the two eqations , you get 3x =12, hence x=4, substitute value in any eq to find y. 2x4 +y =9 , hence y=1.

do the same for the rest

Answer 6

substitution is the easiest unless you are asked to use any other method

1. x=3+y
2(3+y)+7=9
6+2y+7=9
2y=-4
y=-2
x=3-2
x=1

2. y=5+3x
2x+5(5+3x)=8
2x+25+15x=8
17x=-17
x=-1
y=5-3
y=2

3.x=5-3y
2(5-3y)-y=5
10-6y-y=5
-7y=-5
y=5/7
x=5-15/7

do the rest yourself