solve simultaneously....?

4x - 2y = - 6 (1)

5x + 3y = 20 (2)

<=> operation : multiply (1) by 3 and (2) by 2

12x - 6y = - 18 (1 ')

10x + 6y = 40 (2 ')

<=> operation : (1 ') + (2 ') => elimination of " y "

22x = 22

<=>

x = 1

and

4x - 2y = - 6

<=>

4 - 2y = - 6

<=>

2y = 4 + 6

<=>

2y = 10

<=>

y = 5

*******

Answer :

x = 1

and

y = 5

*******

cos(40 5) = (?2)/2 cos(30) = (?3)/2 sin(40 5) = (?2)/2 sin(30) = a million/2 5 + a*cos(40 5) - b*cos(30) = 0.5 + a(?2)/2 - b(?3)/2 = 0 10 + a?2 - b?3 = 0 a?2 - b?3 = -10 enable this be Eqn1. -a*sin(40 5) + b*sin(30) = 0 -a(?2)/2 + b/2 = 0 -a?2 + b = 0 enable this be Eqn2. Eqn1 + Eqn2 b - b?3 = -10 b(a million - ?3) = -10 b = -10/(a million - ?3) b ? 13.6603 -a?2 + b = 0 -a?2 + 13.6603 ? 0 -a?2 ? -13.6603 a ? 13.6603/?2 a ? 9.6593

12x - 6y = - 18
10x + 6y = 40-----ADD

22x = 22
x = 1

5 + 3y = 20
y = 5

x = 1 , y = 5

Here is the workout.

Solve for one variable (doesn't matter which one you pick) from the top equation (again, doesn't matter which one you pick, I pick the first because the coefficients are smaller :-)).

Solve for y 2y=4x+6 Divide both sides by 2 y=2x+3

Substitute in the second one

5x+3(2x+3)=20 Open up the parenthesis

5x+6x+9=20

11x=11

x=1

Substitute for x in y=2x+3

y=2+3=5

So the solutions are x=1 y=5

Now the check

Subsitute for x and y in the first equation

4-10=-6 -6=-6 Check!

Substitute in the second equation

5+15=20 20=20 Check!

This method works better than looking or guessing ......

4x - 2y = -6 (solve by using elimination)
5x + 3y = 20

4x - 2y = -6
2x - y = -3
3(2x - y) = 3(-3)
6x - 3y = -9

....5x + 3y = 20
+) 6x .- 3y = .-9
--------------------------------
..11x ....... = 11

11x = 11
x = 11/11
x = 1

4x - 2y = -6
4(1) - 2y = -6
4 - 2y = -6
2y = 4 - (-6)
2y = 10
y = 10/2
y = 5

∴ x = 1 , y = 5

Your proposal is a correct solution, which you check by substituting those values into each equation.

Your method should be to exclude/negate one of the variables. When you solve "simultaneous equations" you work with both equations at the same time (simultaneously) by adding them together, so to speak. You can multiply each equation by positive or negative values without changing the equations, right.... you're just creating new multiples. But as always, what ever you do to one side of an equation, you must do to the other side. So.....

Let's get rid of y...
3(4x-2y)=-6*3 12x-6y=-18
2(5x+3y)=20*2 10x+6y=40

Ahh... when you add the two equations together, the y term goes away and you're left with 22x=22 and x=1. Now substitute x=1, into one of your original equations, and solve for y
4*1-2y=-6; -2y=-6-4; -2y=-10 y=5. Check to make sure it works for both equations and you're finished.

there's 2 ways to do this:
1. substitution: solve for y alone
4x - 2y = -6
-2y = -6 - 4x
y = 2x + 3
place that into the other equation
5x + 3(2x + 3) = 20
5x + 6x + 9 = 20
11x + 9 = 20
11x = 11
x = 1
plug that back into the original equation
4(1) - 2y = -6
4 - 2y = -6
-2y = -10
y = 5

2. elimination: make either the x's or y's equal to each other but one negative
3(4x - 2y = -6) = 12x - 6y = -18
2(5x + 3y = 20) = 10x+6y = 40
then solve like a normal addition equation
12x + 10x = 22x
-6y + 6y = 0y
-18 + 40 = 22
22x = 22
x = 1
plug x back in to either equation to solve for y
4(1) - 2y = -6
4 - 2y = -6
-2y = -10
y = 5

4x - 2y = -6 . . . . (1)
5x + 3y = 20 . . . (2)

(1) × 3: 12x - 6y = -18 . . . (3)
(2) × 2: 10x + 6y = 40 . . . (4)

(3) + (4):
22x = 22
x = 1

Substitute x = 1 into (1),
4(1) - 2y = -6
4 - 2y = -6
-2y = -10
y = 5

Done

4x-2y=-6
5x+3y=20

3(4x-2y=-6)
2(5x+3y=20)

12x-6y=-18
10x+6y=40
------------------
22x=22 x=1

12-6y=-18
-6y+12-12=-18-12
-6y=-30
y=5

4(1)-2(5)=-6
4-10=-6

x=1 & y=5
4(1)-2(5)=-6
4-10=-6 or -10 + 4=-6

5(1)+3(5)=20
5+15=20