How to solve for x and y? Please help me on how to solve htis problems?

Question

2x-3y=17
3x+y=31

y=2x-1
4x-5y=8

Answer 1

Problem #1 (my math was off at first but I couldn't stop until I got it right!)

2x -3y =17
3x+y = 31

I solved for y using the second equation by subtracting both sides by 3x.

y= 31-3x

Next I substituted y in the 1st equation.

2x -3y =17
2x - 3(31-3x) = 17
2x - 93 +9x = 17
2x +9x = 17 + 93
11x = 110
x = 110/11 = 10

Then I plugged x into the y equation to solve for y.

y = 31-3x
y = 31-3(10)
y= 31-30
y=1

Problem #2

y= 2x-1
4x -5y =8

Since you already have y just plug y in the 2nd equation to solve for x.

4x - 5y = 8
4x - 5(2x-1) = 8
4x + (-5)(2x-1) = 8
4x-10x+5=8
4x -10x = 8-5
-6x=3
x=3/-6 or -1/2

Once you found x, plug it in the y equation to solve for y.

y = 2x-1
y=2(-1/2) - 1
y= (-2/2) - 1
y= -1-1
y=-2

Answer 2

1.) Simultaneously solving the equations, (2x + y = 3)x 6 (5x + 6y = 4)x 1 12x + 6y = 18 5x + 6y = 4 ____________ 7x = 14 x = 2 ---- (I) Subs x in any one of the original equations to get y: 2(2) + y = 3 4 + y = 3 y = -1 ----- (II) x = 2 y = -1 (Answers) 2.) It's a quadratic equation : x² + 3x - 18 = 0 Factorise it: x² + 6x - 3x -18 = 0 x (x+6) - 3 (x+6) = 0 (x+6)(x-3) = 0 Thus, x = -6 or 3 3) Again a quadratic: x² + 9x + 14 = 0 Factorising, x² + 7x + 2x + 14 = 0 x (x + 7) + 2 (x + 7) = 0 (x+2)(x+7) = 0 Thus, x = -2, -7 (Answers) 4) Again a quadratic equation: x² - 8x + 12 = 0 x² - 6x - 2x + 12 = 0 x (x-6) - 2 (x-6) = 0 (x-2)(x-6) = 0 ---> x = 2, 6 (Answers)

Answer 3

Use the elimination method. For example, eliminate the x's:

2x-3y=17 [ multiply all terms by 3, to make the x's = 6x]
3x+y=31 [multiply all terms by 2, to make the x's = 6x]

6x-9y=51 [eqn 1]
6x-2y=62 [eqn 2]

now take [1] from [2]

7y = 11 [ the 6x's cancelled out, and -2 subtract -9 is 7, 62 -51 = 11]

7y=11 [now divide both sides by 7]
y = 11/7

to find x, sub y=11/7 into eqn [2] (it doesnt matter which eqn you choose)

6x-2y=62
6x -2 * (11/7) = 62
6x -1.8571428571428571428571428571429 = 62 [now add 1.8571428571428571428571428571429 to both sides]
6x = 62 + 1.8571428571428571428571428571429
6x = 63.857 (rounded to 4 decimal places)
divide both sides by 6
x = 10.643

hope that helps, would be great if someone else could check my answers, just in case.

Answer 4

on first one, solve for y in the second equation to get y=31-3x
then plug 31-3x in for y in the first equation to get
2x-3(31-3x)=17
once you get an answer for this, plug the answer (x value) into the 2nd equation to get 3(answer to first)+y=31. Use this to get the y value

repeat the same process for #2 except you already have y=2x-1 and dont need to solve for y like you did in the first equation

this is called substitution

Answer 5

sol1- multiply eq2 by 3 then add the new eq from eq1
(2x - 3y =17)+ (9x + 3y = 91)
u will get
11x = 108
x= 108/11
put the value of x in eq 1 we get
2* 108/11 -3y = 17
-3y = 17 - 216/11
3y = 216/11-17
y= 29/33

solve the next problem similarly

Answer 6

hey this question number one can be solved by this.
First give number to equation like
2x-3y=17...........(1)
3x+y=31............(2)
Now try to solve both equations .you know when putting both equations togther and change the signs like
2x-3y=17
3x+y=31 but look sign will be now change then equation 2 will be
-3x-y=-31 now solve


2x-3y=17
-3x-y=-31
by solving we get -x-4y=-14 taking y to other side
-x=-14+4y now multiply by negative sign on both sides.
x=14-4y now put the value of x in first equation and get the y value then after getting y value put into 2nd equation then you will get x value
Like this you can solve for question number 2.

Answer 7

to solve for y set x=0

to solve for x set y=0

Answer 8

http://library.thinkquest.org/20991/alg2/systems.html

or find another math website by googling How to solve for x and y