Algebra Question, 2 system equations.?

2x + 5y = 10
3x - 2y = -4

==> 3 (2x + 5y = 10)
==> 2 (3x - 2y = -4)

6x + 15y = 30
6x - 4y = - 8
- + +

19y = 38

y = 38/19

y = 2

2x + 5y = 10
5y = 10 - 2x
y = (10 - 2x)/5

3x - 2y = - 4
2y = 3x + 4
y = (3x + 4)/2

Value of x:
2(10 - 2x) = 5(3x + 4)
20 - 4x = 15x + 20
19x = 0
x = 0

Value of y:
= (10 - 2[0])/5
= (10 - 0)/5
= 10/5 or 2

Answer: y = 2

Proof (2nd equation):
3(0) - 2(2) = - 4
0 - 4 = - 4
- 4 = - 4

y = x²-2x+a million a=a million b=-2 c=a million b²-4ac = 4-4 = 0 y = a million/4 x² + a million/2 a=a million/4 b=0 c=a million/2 b²-4ac = 0-a million/2 = -a million/2 2. those are the two purposes, which could have endless form of ideas. If y = 0 then: 0 = x² - 4 = (x+2)(x-2) There are 2 ideas, x = ±2 0 = 3x² + 6x + 2; locate the discriminant b²-4ac = 36 - 24 = 12 this is a favorable variety, subsequently there are 2 ideas (if the discriminant is 0 there is one answer, no rely if this is detrimental there are no longer any ideas

4x + 10y = 20
15x - 10y = - 20----ADD

19x = 0
x = 0

5y = 10
y = 2

x = 0 , y = 2

y = -2/5*x+2

y = 3/2*x+2

answer: y=2.
2x+5y=10
2•0+5•2=10
0+10=10

2x + 5y = 10

3x-2y =-4
1] y =10/5
=2
2]y=-4/2
=2

3x=2y-4
x=(2y-4)/3

2((2y-4)/3)+5y=10
2(2y-4)/3+5y=10 (*3)
2(2y-4)+5y*3=10*3
2*2y-2*4+15y=30
4y-8+15y=30
4y+15y=30+8
19y=38
y=38/19
y=2

x=(2y-4)/3
x=(2*2-4)/3
x=0

Check-up:

2*0+5*2=0+10=10

3*0-2*2=0-4=-4

2x + 5y = 10 ----- (1)
3x - 2y = -4 ----- (2)

(1)*3 - (2)*2:
6x +15y - 6x + 4y = 30 + 8
19y = 38
y = 2

Hence, y = 2.

2x + 5y = 10 (solve by using elimination)
3x - 2y = -4

2x + 5y = 10
-3(2x + 5y) = -3(10)
-6x - 15y = -30

3x - 2y = -4
2(3x - 2y) = 2(-4)
6x - 4y = -8

...-6x - 15y = -30
+) 6x - 4y = -8 (addition)
--------------------------------
-19y = -38

-19y = -38
y = -38/-19
y = 2

∴ the value of y is 2.