<span>Multiply one of the equations so that both equations share a common complementary coefficient.
In order to solve using the elimination method, you need to have a matching coefficient that will cancel out a variable when you add the equations together. For the 2 equations given, you have a huge number of choices. I'll just mention a few of them.
You can multiply the 1st equation by -2/5 to allow cancelling the a term.
You can multiply the 1st equation by 5/3 to allow cancelling the b term.
You can multiply the 2nd equation by -2.5 to allow cancelling the a term.
You can multiply the 2nd equation by 3/5 to allow cancelling the b term.
You can even multiply both equations.
For instance, multiply the 1st equation by 5 and the second by 3. And in fact, let's do that.
5a + 3b = –9
2a – 5b = –16
5*(5a + 3b = -9) = 25a + 15b = -45
3*(2a - 5b = -16) = 6a - 15b = -48
Then add the equations
25a + 15b = -45
6a - 15b = -48
=
31a = -93
a = -3
And then plug in the discovered value of a into one of the original equations and solve for b.</span>
Answer:
-60
Step-by-step explanation:
If the worker is descending, he is going down. The change in his position could be marked on a graph by translating the original point down by 60 units. Aka -60 units.
B.) 8(x+4) is the best choice
To solve for m we want to get m by itself on one side of the equals sign, with everything else on the other side. First we subtract b from both sides, to get it out of the right side of the equation:
y - b = mx + b - b
y - b = mx
Now we divide both sides by x, to get m completely by itself:
(y-b)/x = (mx)/x
(y-b)/x = m
So m is equal to (y-b)/x.
So it is asking you to group like term so
x terms can be grouped/added/subtracted to other x terms, but not to x^2 or x^3 terms
x^2 terms to x^2 and so on so
1. 9-3k+5k=
9+(5k-3k)=
9+2k
2. k^2+2k+4k=
k^2+(2k+4k)=
k^2+6k=
