Step-by-step explanation:
x=4 and y= 1 because of 4+1=5 and add 2y to both sides
x-2y+2y=2+ 2y
x=2y+2
4-2=2
[tex]\left \{ {{x+y=5} \atop {x-2y=2}} \right.[/tex]
Here we have 2 equations in one system.
System of Equations/Simultaneous Equations are for finding the intercepts of both graphs.
We can solve these Simultaneous Equations in many ways. However, I'll be solving by substitution.
To solve by using substitution, we need to move either x or y to the another side. I'll move x term to another side for the first equation. (We'll substitute the first equation in the second.)
[tex]\left \{ {{y=-x+5} \atop {x-2y=2}} \right.[/tex]
Now we substitute y = -x+5 in x-2y = 2
[tex]x-2(-x+5)=2\\[/tex]
Then distribute -2 in (-x+5) | Note that multiplying the negative will get positive. |
[tex]x-2(-x+5)=2\\x+2x-10=2\\3x=2+10\\3x=12[/tex]
We should get this. When we get this, we move 3 to divide 12 (Opposite when moving to another side so from plus to minus and from multiply to divide.)
[tex]x=4[/tex]
But we are not done yet. We have to find the y-value. After all they are two variables equations.
If we want to find the y-value then we have to substitute x = 4 in any given equations. I'd suggest to substitute in the equation that doesn't have a coefficient with high value.
So I'll be substituting in x + y = 5 instead.
Substitute x = 4 in x + y = 5
[tex]4+y=5\\y=5-4\\y=1[/tex]
Thus, we get both x and y value. Answer in either coordinate point or writing form.
Therefore, the answer is
x = 4 when y = 1 or (4,1)
Both works. Hope this helps.
