How do i do theese types of math problems (explain please) algebra house ill best answer u 10points

x - y = 5
2x + y = 4

x - y = 5
2x + y = 4
-------------------
3x = 9 {added the two equations together}
x = 3 {divided both sides by 3}

2x + y = 4 {bottom original equation}
2(3) + y = 4 {substituted into bottom equation}
6 + y = 4 {multiplied 2 by 3}
y = -2 {subtracted 6 from both sides}

x = 3 and y = -2

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You have to put one of the equations in the terms of the other.

This problem:

x - y = 5
2x + y = 4

Therefore, y = x - 5

Now take that y and insert it into the second equation, thusly:

2x + y = 4

2x + (x-5) = 4

2x + x - 5 = 4

3x - 5 = 4

3x = 9

x = 3

Now that you know x, pick an equation and plug it in to solve for y.

3 - y = 4

-y = 1

y = -1

x-y=5
2x+y=4

multiply (-2) times the top in order to subtract it from the bottom and find for y.

that gives 3y=-6
so y=-2

then plug in -2 for y to get x-(-2)=5

subtraction of 2 gives x=3

so the final answer is x=3 and y=-2

First method: substitution.
From x-y=5, we can say x=y+5.
Plug in for x in the next equation, 2(y+5)+y=4
2y+10+y = 4
3y=-6
y=-2
so x=3

Second method: combining equations
Add them together.
x-y=5
2x+y=4
______
3x=9,
x=3
so y=-2

x - y + 2x + y = 5 + 4
3x = 9
x = 3

3 - y = 5
y = - 2

Answer: x = 3, y = - 2

Check—2nd equation:
2(3) - 2 = 4
6 - 2 = 4
4 = 4