Solve System for 2x-3y=2, 3x-y=10?? plz help out?

Answer 1

#1: 2x-3y=2
#2: 3x-y=10

Solve #2 for y.
3x - y = 10
3x = 10 + y
3x - 10 = y

Plug that into #1 and solve for x.
2x - 3y = 2
2x - 3(3x-10) = 2
2x - 9x + 30 = 2
-7x = -28
x = 4

Plug that back into #2 and find y.
y = 3x - 10
y = 3(4) - 10 = 12 - 10 = 2

Answer 2

Use the elimination method: 2x-3y=2 3x-y=10 Times either the top or the bottom equation by one number to subtract the letter out. For example, I will times the bottom equation by a -3 which will make the equation look like this: 2x-3y=2 -9x+3y=-30 Notice how the y's will now subtract out: 2x-3y=2 -9x+3y=-30 ========= -7x = -28 x = 4 Now go to ONE of the ORIGINAL equations and plug in 4 for x and solve for y. I will use the equation 2x-3y=2. 2(4) - 3y = 2 8 - 3y = 2 -3y = -6 y = 2. Hope this helps :)!

Answer 3

Multiply the second equation by -3 and add the resulting equations.
2x-3y=2
-9x+3y=-30
You get -7x=-28 which gives x=4
Substitute this into the first equation to get 8-3y=2 or 3y=6 so y=2.
Check 2(4)-3(2)=2 and 3(4)-2=10

Answer 4

*ISolate "x" or "y" in the either equation. Let's isolate "x" in the 1st equation.

First: add "3y" to both sides (when you move a term to the opposite side, always use the opposite sign).

(1) 2x-3y+3y = 2+3y
2x = 2+3y (divide both sides by 2).
2x/2 = 2/2 + 3y/2
x = 1 + 3y/2

Sec: substitute "1+3y/2" with "x" in the 2nd equation.

(2) 3(1+3y/2) - y = 10
3(1)+3(3y/2) - y = 10
3+9y/2 - y = 10
3 + 7y/2 = 10 (subtract 3 from both sides).
3 -3 + 7y/2 = 10-3
7y/2 = 10-3
7y/2 = 7 (multiply 2/7 with both sides).
(2/7)(7y/2) = (2/7)(7)
y = 14/7
y = 2

Fourth: substitute 2 with "y" in the 1st equation.

(1). 2x-3y = 2
2x - 3(2) = 2
2x - 6 = 2 (add 6 with both sides).
2x - 6+6 = 2+6
2x = 2+6
2x = 8 (divide both sides by 2).
2x/2 = 8/2
x = 8/2
x = 4

Solution set (4, 2)

Answer 5

you need simultaneous equations.
the idea is that you use both of them to find out what x and y values they ahve in common
so
2x-3y=2 eqn 1
3x-y=10 eqn 2

multiply equation 1 by 3 and eqn 2 by 2 so that the x coefficients cancel out.
6x-9y=6 NEW eqn 1
6x-2y=20 NEW eqn 2
solve to find y
-7y=-14
y=2

then sub in y=2 into eqn 1
and hey presto
2x-3(2)=2
x=4

hope that helped ;]

Answer 6

2x-3y=2 (i)
3x-y=10 (ii)
first multiply (ii) by 3 giving new (ii) 9x-3y=30
now subtract (i) from new (ii)
9x-3y=30
2x-3y=2
giving 7x=28
so x=4
now substitute in original (ii)
12-y=10
so y=2

so x=4; y=2

check in the original (i)
2*4-3*2
8-6
=2

Answer 7

2x-3y=2 and 3x-y=10 use one equation and equate one variable then substitute to the other equation if u dont understand follow my solution

2x-3y=2 and 3x-y=10

using the second equation

y=3x-10 so substitute in the first equation

2x-3(3x-10)=2

2x-9x+30=2

-7x=-28

x=4

then y= 3x-10

y=3(4)-10

y=12-10

y=2

the solution is (4,2)

or use this

2x-3y=2 and 3x-y=10

2x -3y -2=0 3x -y -10=0
a1 b1 c1 -- a2 b2 c2

x= [c1(b2)-c2(b1)]/[a2(b1)-(a1)(b2)]

y=[a1(c2)-a2(c1)]/[a2(b1)-a1(b2)]

x=[-2(-1) - (-10(-3))] / [3(-3) - 2(-1)]

x= [2-30]/[-9+2]

x=-28/-7

x=4

and

y=[2(10) - (3)(-2)]/[-3(3) - (2)(-1)]

y=[-20+6 ]/[-9+2]

y= -14/-7

y=2

(4,2) same answer sorry for the late update


2x-3y-2 and 3x-y-10

Answer 8

2x - 3y = 2
- 9x + 3y = - 30-----ADD
- 7x = - 28
x = 4
12 - y = 10
y = 2
SOLUTION is x = 4 , y = 2

Answer 9

This is pretty easy, if I were you, i would use elimination

2x-3y=2------->3------->6x-9y= 6
3x-y=10------>-2------->-6x+2y = -20
7y=-14
y=-2
plug that in
3x-(-2)=10
3x+4=10
3x=6
x=2, y=-2
(2,-2)

Answer 10

use your search facility to find dr math, its a good site and explains and answers it all for you.