SOLUTION: 2x+3y=1 5x+3y=16 Solve, state if inconsistent or dependent 3y=2x+1 divide both sides by 3y= y=2x+1/3 y=2x/3+1/3 confused on the LCM...6? I believe once I reduce the top I

Question 126654This question is from textbook Beginning Algebra
: 2x+3y=1
5x+3y=16 Solve, state if inconsistent or dependent
3y=2x+1
divide both sides by 3y=
y=2x+1/3
y=2x/3+1/3
confused on the LCM...6?
I believe once I reduce the top I then add the y into the bottom equation of
5x+(3)( y )x+(3)(y)=16
I can not go any further because I am confused on the LCM of the first equation This question is from textbook Beginning Algebra

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
You got off to a good start by solving one of the equations for y, except for a sign error, but then your process seems to go astray.

I presume since you began by solving one of the equations for y, that you want to solve this system by the substitution method. I'm not sure that I would have chosen substitution given the fractional coefficients once you solved the first equation for y, but let's continue anyway.

is your expression of the first equation in terms of y, but when you moved the 2x to the right side of the equation, you forgot the minus sign -- you had to add -2x to both sides to move the term. So, your equation should be

In order to proceed with the substitution method, you need to replace y in the second equation by something you know to be equal to y, namely

The rest is just manipulation:

Now that we have a value for x, we can substitute this into either of the equations to calculate y, and we can use the equation already solved for y:

And your solution set contains the single ordered pair, (5,-3). Since there is only one element in the solution set, the system is consistent and independent (also, and probably more accurately, referred to as 'uniquely determined')

My comment about the substitution method being less than optimum certainly begs the question of how I would proceed otherwise.

I would use the elimination method:

Eq 1:
Eq 2:

We want to find a constant value or values that we can use to multiply either or both equations so that we end up with the coefficients on one of the variables being additive inverses of each other. Here, we can multiply either of the equations by -1 resulting in a -3 coefficient on y in one of the equations and a 3 coefficient on y in the other.

Multiplying Eq. 1 by -1, we get:
Eq 1:
Eq 2:

Now, we add the two equations, term-by-term, to get:

And solve for x:

Now you could, if you were so inclined, go back to the original two equations, multiply the first one by -5 and the second one by 2 giving you -10 and 10 as the coefficients on x. Then the addition of the equations would eliminate x, allowing you to solve for y. But that's more work than I care to do. Just substitute the value for x into either equation and solve for y, just like we did to complete the substitution method. As you might expect, we will get the same result as before, namely y = -3.

Again, the solution set contains the single ordered pair (5,-3).

Just because some of us are visually oriented, here's a graph of the system to give us a nice warm fuzzy feeling that we have solved the problem correctly.

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