Solve by Elimination Show your work?

Multiply first equation by 3 and add to second equation:

6y + 3x = 9
4y - 3x = 1
- - - - - - - - - -
10y + 0x = 10
10y = 10
y = 1 <====

from first equation
2*1 + x = 3
x = 1 <====

2y + x = 3
4y - 3x = 1

2y + x = 3
2(2y + x) = 2(3)
4y + 2x = 6

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

5x = 5
x = 5/5
x = 1

2y + x = 3
2y + 1 = 3
2y = 3 - 1
y = 2/2
y = 1

∴ x = 1 , y = 1

6y + 3x = 9
4y - 3x = 1-----ADD

10y = 10
y = 1

2 + x = 3
x = 1

x = 1 , y = 1

multiply the top equation by 3 then add the two equation therefore eliminating the x's

3(2y + x = 3)---> 6y + 3x =9

6y +3x =9
4y - 3x = 1
------------------
10y = 10
y = 1

replace y in one of the original equation to find your x

2(1) + x =3---> 2 + x = 3
x=1
These two line intersection at point (1,1)

Solve by elimination

2y + x = 3 ===> Eq. 1

4y - 3x = 1 ====> Eq. 2

Multiply Eq. 1 by 3

3( 2y + x = 3) =

6y + 3x = 9 ===> Eq. 3

Add Eq. 2 and Eq. 3

4y - 3x = 1 ===> Eq. 2
+ (6y + 3x = 9) ===> Eq. 3
-------------------------------------
10 y = 10

y = 10/10

y = 1

Substitute the value of y into Eq. 1 ( 2 or 3) to find x

2(1) + x = 3

2 + x = 3

x = 3 - 2

x = 1
===========
To check
x = 1, y = 1

2y + x = 3 ==> Eq. 1
2(1) + 1 = 3

2 + 1 = 3

3 = 3

4y - 3x = 1 ==> Eq. 2
4(1) - 3(1) = 1

4 - 3 = 1
1 = 1

call 2y + x = 3 number [1]
and 4y - 3x = 1 number [2]

multiply [1] by 2:
4y + 2x = 6 [3]

[3] - [2]

4y - 4y + 2x --3x = 6 -1
5x = 5
x = 1 [4]

[4] ---> [1]
2y + 1 = 3
2y = 2
y = 1

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

-4y - 2x = -6
4y - 3x = 1
0 - 5x = -5
x = 1
2y + 1 = 3
2y = 2
y = 1

multiply the 1st eqn by 3. it would become
6y + 3x = 9
4y - 3x = 1 (2nd eqn)
add the two
10 y +0x = 10
this gives y=1
put value in first eqn
2*1 +x = 3
thus x = 1

x = 3 - 2y
4y - 3(3 - 2y) = 4y - 9 + 6y = 10y - 9 = 1
y = 1
2(1) + x = 3
x = 1

Is that elimination?