It depends on what variable you are tying to solve for first. Say you are trying to solve for x first and then y on the first problem you wrote.
In substitution you solve one of the equations for example with
6x+2y=-10
2x+2y=-10
you solve 2x+2y=-10 for x
2x+2y=-10
-2y = -2y (what you do to one side of the = you do to the other)
2x=-10-2y (to get the variable by its self you divide the # and the variable)
/2=/2 (-10/2=-5 and -2y/2= -y or -1y, they are the same either way)
x=-5-y
now you put that in your original equation that you didn't solve for:
6(-5-y)+2y=-10 solve for that
-30-6y+2y=-10 combine like terms
-30-4y=-10 get the y alone and to do this you first get the -30 away from it
+30=+30
-4y=20 divide the -4 from each side
/-4=/-4 (20/-4=-5)
y=-5
now the equation you previously solved for x can be solved for y.
x=-5-y
x=-5-(-5) a minus parenthesis negative -(- gives you a positive
-5+5=0
x=0
and now we have solved the problem. x=0 and y=-5
2d +9= 19
The solution is d =
Hello there, first we can place our equation, which you have stated
, this algebra, in algebra we want to find the exact value of the variable, or have the variable on one side of the equation.
First we can subtract 
with both sides of the equation giving us the result of 
, this can also be represented as 
, now just divide both sides of the equation by 
, our answer is 
.
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Answer:
$0.23
Step-by-step explanation:
Theres 16 ounces in a pound. 16-1.7=14.3
divide $3.31 by 14.3
0.23
Answer:
hold on i'm abt to answer it
Step-by-step explanation:
Answer:
190% of 66= 125.4
rounded to the nearest tenth:130
