From the equation of a line;
y=mx+c
m=2
y=2x+c
Replacing for;
y=1
x=3
1=2*3+c
1=6+c
c=1-6
c=-5
<span>y=2x-5</span>
Answer:
IT is not
Step-by-step explanation:
Let's replace. If it's a solution, it means that BOTH equations has to be true.
Now, the second equations seems good - 
indeed.
First alas isn't. RHS is 142, which is totally not the same as -10.
Answer:
x,y =2,1
Step-by-step explanation:
3x+4y=10 ................equ1
x-y=1 ................equ2
This system of equations forms a quadratic equation.
Lets use the substitution method in solving this set of equations
from equ 2; x=y+1
substitute x=y+1 into equ 1
3(y+1) +4y=10
3y+3+4y=10
7y=10-3
7y=7
y=7/7
y=1
but x=y+1
substitute the value of y into this equation to get x
x= 1+1
x=2
x,y=2,1
Answer:
19y-9
Step-by-step explanation:
apply distributive property
-3(-4y)-3 x 3 + 7y
12y - 3 x 3 + 7y
12y - 9 + 7y
add 12y to 7y
19y-9
P= 2l+2w
P-2l= 2w
1/2P -l= w
Rearrange into standard form.
w= 1/2P -l
