Question

which of the following is a solution to this system of equations: 5x + 3y = 4 3x +y =1

Answer 1

Well, you can either solve the solution of equations by elimination or substitution or whatever other method you might know, or you can try out the answer choices and see which one satisfies both equations. I suggest that you solve the system of equations, because you don't always have the luxury of "multiple guess" questions, and if you practice doing it, you'll be faster at doing that then testing a bunch of possible answers to see which one works.

Answer 2

If you need instruction on how to solve such a system, say so, and I'll teach you.

Answer 3

i need instructions---------

Answer 4

I was hoping you would say that :-)

\[5x + 3y = 4\]\[3x+y=1\]

There are two main approaches that people who are just getting started with systems of equations tend to use: substitution, and elimination. I'll describe both, because you should know how to do both, although for any given problem, one may be appreciably easier than the other.

Answer 5

it was 5x + 3y = 5

Answer 6

When you have a system of two equations in two unknowns like this, each equation describes a line. One way to solve the system is to graph each line, and find the point(s) at which they intersection, if any. There are 3 basic cases:

1) they intersect at a point. That point is the solution.
2) they are parallel, and do not intersect. There are no solutions.
3) the lines are identical, and every point on the line is a solution (infinitely many).

Answer 7

Okay, we'll use that value when we do the problem. Thanks for noticing and pointing it out.

In substitution, we take one of the equations, and solve it for one variable in terms of the other(s). This is convenient if one of the equations (like \(3x+y=1\) here) is already almost solved for us, and only minimal algebra is required. We solve the equation:
\[3x+y=1\]Subtract \(3x\) from each side:
\[3x-3x+y=1-3x\]
What do you get if you simplify that by collecting like terms?

Answer 8

y = -2

Answer 9

@kpotter this is an interactive lesson; you need to take part in a timely fashion.

Answer 10

3x + y =1
3x -3x =0 1- -3x = -2x
then you bring down your y=-2

Answer 11

No, you just get
\[3x-3x+y = 1-3x\]\[\cancel{3x}\cancel{-3x}+y = 1-3x\]\[y = 1-3x\]That's all that you can do for the moment.

Answer 12

Now, you take the other equation, \( 5x + 3y = 5\), and everywhere that \(y\) appears in that equation, you replace it with \((1-3x)\) giving rise to the name substitution method.

What do you get if you do that?

Answer 13

Use the distributive property to get rid of the parentheses in your answer.

Answer 14

5x + 3(1-3)= 5
5x + 6 = 5
5x -5 +6 5- -5
i dont know what to do from here

Answer 15

@whpalmer4

Answer 16

i get 5

Answer 17

No, you didn't follow the directions correctly. I didn't say substitute (1-3), I said (1-3x).

Answer 18

i think i got it now and if im solving a system by elimination what would be my first stsep to solving -6x +9y = -72
6x +2y = 6

Answer 19

Look, it's simple:

\[5x+3y=5\]Now replace \(y = 1-3x\):
\[5x+3(1-3x) = 5\]\[5x + 3*1 -3*3x = 5\]\[5x + 3 -9x = 5\]\[-4x + 3 = 5\]\[-4x = 5-3\]\[-4x=2\]\[x = -1/2\]

Answer 20

Now you put that value into the other formula:\[y = 1-3x\]\[y = 1-3(-\frac{1}{2})\]\[y = 1+\frac{3}{2}\]\[y = \frac{5}{2}\]So the solution to this system is \((\large -\frac{1}{2},\frac{5}{2})\)

And we can see that on the graph I've attached:

Answer 21

So I think this other system is a good example of where you would use elimination, not substitution.

In elimination, we try to arrange for the equations to have the same coefficients (except opposite in sign) for one of the variables.

\[-6x +9y = -72\]\[ 6x +2y = 6\]

Here we have \(6x\) and \(-6x\). What happens if you add \(6x\) and \(-6x\)?

Answer 22

it will equal zero

Answer 23

it will be e-liminated

Answer 24

Right. That's what we want.

\[-6x+9y=-72\]\[6x+2y=6\]We add those two equations:

\[-6x+6x + 9y + 2y =-72 + 6\]
What does that simplify to?

Answer 25

my mom helped me i have another question---