y = 2x + 2
y = 1/2x + 2
If I were to solve this system of equation through subsitution, how would the setup look like.
But im not sure if it is able to be solved with subsitution, Because the first part of the question was what system of equation could this equation be solved with, Multiplication, subsitution, or graphing. And I said subsitution, Im not sure if I was correct or not.
The questions revolve around the very top equation.
Please help.
y = 1/2x + 2
If I were to solve this system of equation through subsitution, how would the setup look like.
But im not sure if it is able to be solved with subsitution, Because the first part of the question was what system of equation could this equation be solved with, Multiplication, subsitution, or graphing. And I said subsitution, Im not sure if I was correct or not.
The questions revolve around the very top equation.
Please help.
-
Any pair of coplanar linear equations can be solved as long they don't have the same slope. (If they have the same slope they are parallel and don't intersect, unless they actually describe the same line, in which case they have infinite solutions.)
y = 2x + 2
y = 1/2x + 2
We have to adjust one of these equations so that one of the variables (y or x) exactly cancels out the corresponding variable in the other equation. The easiest way to do that would be to multiply one of the equations by -1. I'll multiply the top equation by -1, although it would work just as well if you chose the bottom one.
(-1)(y = 2x + 2) ===> -y = -2x - 2
y = 1/2x + 2
Now combine the equations, adding the left-hand terms to the left-hand terms and the right-hand terms to the right-hand terms:
y - y = -2x + 1/2x - 2 + 2
And cancel where necessary. y and -y will cancel out the left side completely, leaving zero.
0 = - 3x/2
Solving for x is easy:
x = 0
So now we can plug in zero for x in either one of the equations to solve for y:
y = 2(0) + 2
y = 2
Or....
y = 1/2(0) + 2
y = 2
Either way you look at it, y = 2. The solution is therefore (0, 2). I hope this helps. Good luck!
y = 2x + 2
y = 1/2x + 2
We have to adjust one of these equations so that one of the variables (y or x) exactly cancels out the corresponding variable in the other equation. The easiest way to do that would be to multiply one of the equations by -1. I'll multiply the top equation by -1, although it would work just as well if you chose the bottom one.
(-1)(y = 2x + 2) ===> -y = -2x - 2
y = 1/2x + 2
Now combine the equations, adding the left-hand terms to the left-hand terms and the right-hand terms to the right-hand terms:
y - y = -2x + 1/2x - 2 + 2
And cancel where necessary. y and -y will cancel out the left side completely, leaving zero.
0 = - 3x/2
Solving for x is easy:
x = 0
So now we can plug in zero for x in either one of the equations to solve for y:
y = 2(0) + 2
y = 2
Or....
y = 1/2(0) + 2
y = 2
Either way you look at it, y = 2. The solution is therefore (0, 2). I hope this helps. Good luck!
-
if you used substituion then y=y
2x+2=1/2x+2.
solve for x
3/2(x)=0
x=0.
2(0)+2
y=2.
interception at (0,2)
you can use substitution. You are correct. Your teacher is trying to get you to think.
2x+2=1/2x+2.
solve for x
3/2(x)=0
x=0.
2(0)+2
y=2.
interception at (0,2)
you can use substitution. You are correct. Your teacher is trying to get you to think.
-
it has a solution
y is defined in both terms
so
2x + 2 = 1/2x + 2 /- 0.5x
1.5x+2 = 2 / -2
1.5x = 0 / div by 1.5
x = 0
given x = 0
y = 2*0 +2 1/2*0 + 2
y = 2 y= 2
2=2
solved
y is defined in both terms
so
2x + 2 = 1/2x + 2 /- 0.5x
1.5x+2 = 2 / -2
1.5x = 0 / div by 1.5
x = 0
given x = 0
y = 2*0 +2 1/2*0 + 2
y = 2 y= 2
2=2
solved
-
2x+2=1/2x+2 so 2x=1/2 x and 3x=0 s0 x=0