Solve the system by addition?

Question

Solve the system by addition.5x – 3y = 13 and4x – 3y = 11

Answer 1

5x-3y=13
by adding 3y on both sides
5x-3y+3y=13+3y
5x=13+3y
by dividing both sides by 5
5x/5=(13+3y)/5


Then put the value of x in the other equation:

4x-3y=11
4(13/5+3y/5)-3y=11
4(13/5)+4(3y/5)-3y=11
52/5+12y/5-3y=11
52+12y-15y=11
---------------
5

52-3y=11
-------
5

multiplying both side by 5
......52-3y
5*--------- = 11*5
........5

52-3y=55

adding 3y on both sides
52-3y+3y=55+3y
52=55+3y
substracting 55 from both sides
52-55=55-55+3y
-3=3y
dividing both sides by 3
-3=3y
--...---
3.....3
-1=y


Now put the value of y in the equation

x=13/5+3y/5

x=13/5+3(-1)/5
x=13/5-3/5
x=13-3
.....-----
.......5
x=10
....----
.....5
x=2

So x=2 and y=-1

Answer 2

Solving a system of equations by addition means to literally add the 1st and 2nd equations together.

That means

5x - 3y = 13
+ (4x - 3y = 11)
----------------------

Okay.. so to approach this, try this tip: instead of "adding" let's "subtract" (and yes that's the same thing, think of "adding" a negative number: 5 + (-4) = 1)

So try this..
5x - 3y = 13
- (4x - 3y = 11)
----------------------
= 1x + 0y = 2 (remember that -3 - (-3) = -3 + 3 which is 0 )

so now you have that x = 2.

Now plug x = 2 back into EITHER equation (your choice!)
So for example.. I'll use the 5x - 3y = 13 equation.
Plug x in: 5*(2) - 3y = 13

Using Algebra to solve for y:
-3y = 13 -10
-3y = 3
y= -1

And there you have it. Your "final" answer is x= 2, y = -1.
Or sometimes it may be written as (2,-1) is your solution. Anyway, hope that helped. And let me know if my explanations were confusing.

Answer 3

5x - 3y = 13
4x - 3y = 11

Step 1. Multiply 2nd equation by -1.
5x - 3y = 13
-4x + 3y = -11

Step 2. Add equations together (3y and -3y cancel each other out.)
5x - 3y = 13
-4x + 3y = -11
-------------------
x = 2

Step 3. Substitute x=2 into the first equation.
5(2) - 3y = 13
10 - 3y = 13
-3y = 3
y = -1

4. The solution to your system of equations is (2, -1)

Answer 4

5x - 3y = 13.....A
4x - 3y = 11.....B

Addition will not help as when you add A and B, you get 9x - 6y = 24, which can be solved only by sustitution.

Instead, subtract B from A
(5x - 3y) - (4x - 3y) = 13 - 11
5x - 3y - 4x + 3y = 2
5x - 4x - 3y +3y = 2
x = 2.....C

Put C in any equation to find y. I'll prefer to substitute C in B to get y
4*2 - 3y = 11
8 - 3y = 11
-3y = 11-8
-3y = 3
y = -1

x = 2, y = -1

Answer 5

to use addition you need to multiply one equation by -1 then add them together:
5x - 3y = 13
-4x +3y = -11

add them:
5x - 4x - 3y + 3y = 13 - 11
x = 2

then solve for y:
8 - 3y = 11
-3y = 3
y = -1

Answer 6

5x – 3y = 13
Multiplying the 2nd equation by -1,
-4x + 3y = - 11
The y's drop out and all you have left is x. Find y by substitution, OR
multiply the 1st equation by 4:
20x - 12y = 52
multiply the 2nd (original) equation by -5:
-20x + 15y = -55
This leaves
3y = -3

Answer 7

truly you're meant to characteristic both equations jointly. in case you look how they're coated up, you upload down the columns from proper to left in simple terms like you would do in case you further wide-spread numbers. hence, 6 + 2 = 8 y + -y = 0 and x = x = 2x. you're left with 2x + 0 = 8. The 0 is only a placeholder. you're truly left with 2x = 8. 2x truly means 2 circumstances x. branch undoes multiplication so that you divide each and each area by 2. you're left with x = 4. you could now plug x into both equation to sparkling up for y.

Answer 8

5x=13+3y
x=(13+3y)/5
then
4(13/5+3y/5)-3y=11
52/5+12y/5-3y=11
52/5-3y/5=11
52/5-11=3y/5
-3/5=3y/5
-1=y

x=(13+3y)/5
x=(13-3)/5
x=10/5
x=2