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Topic: Systems of Equations - Please Help?**Question:**
I'm working on a algebraic problem that involves systems of equations. I mostly understand how to solve them, but I still become confused when I try to figure out which variable would be easiest to isolate first.
Here are two of the problems I am having trouble with:
The directions say to 'solve each system of equations.'
3x + 2y = 0
x - y = -4
And...
Ryan and Juan collect baseball cards. Together they have 880 cards. Juan has 125 fewer than twice as many as Ryan. How many cards does each have? Write a system of equations and solve.
If you could give me some helpful tips on how to approach these, that would be great.

July 20, 2019 / By Allyson

Assuming you are doing the substitution method, it's usually easier to isolate the variable with no number in front. Like, the x in equation 2. You could just add y to both sides: x = y - 4 Then replace the other x with (y - 4) 3(y - 4) + 2y = 0 3y - 12 + 2y = 0 5y - 12 = 0 5y = 12 y = 12/5 Then plug that in to get x. On the other, r = Ryan's cards j = Juan's cards r + j = 880 j = 2r - 125 Here, j is already isolated in equation 2 so it's easy; replace the j in eq. 1 with 2r - 125 r + 2r - 125 = 880 3r - 125 = 880 3r = 1005 r = 1005รท3 = 335 then use that to get j

👍 238 | 👎 5

Did you like the answer? 1) x> -1 you just take the -3x and add it to the left side with the 2x, so it's 5x. Then you subtract the 2 from the left and put it to the right, making it -3-2, which is -5. 5x divided by -5 is -1. It's the same thing even as solving it with an equal sign. 2) x= -5/3 y= 10/3 you see how y=-2x? well you take that, and insert it where the y is in y-x=5. So that makes it -2x-x=5. -2x-x is -3x=5. Divided 5 by 3x and you get x=-5/3. Then you place x inside either equations. I chose the second one because it's easier. So then it's y= -2(-5/3). That equals 10/3. You can insert the answers back into the problems to check your answer.

#1 2x +2> -3x-3 5x+2> -3 ( add 3x to both sides) 5x>-5 ( subtract two from both sides) x>-1 ( divide both sides by 5) #2 try replacing y in the first equation with -2x since Y=-2x (-2x) -x=5 now just solve the equation for x don't forget to go back and also solve for y by entering your answer for x

Ok for the first system: Think of 3x + 2y = 0 as equation 1 From x - y = -4 we can derive that x = y - 4, this is equation 2. Now we substitute equation 2 into equation 1: 3(y - 4) + 2y = 0 Multiply the 3 through the bracket: 3y - 12 + 2y = 0 Shift the -12 to the other side and add like terms: 5y = 12 Solve for y: y = 12/5 = 2.4 Substitute this value of y into either of the other two equations (equation 2 would be easiest) and solve for x: x = 2.4 - 4 = -1.6 The reason why it is easiest to isolate for y first in this case is because the x in the second equation has no coefficient, making it easiest to produce an equation with which we can represent x, substitute, and solve for y. For the second problem: Let x represent the number of cards that Juan has Let y represent the number of cards that Ryan has We know that the sum of their cards is 880 so our first equation will be x + y = 880 The second equation is x = 2y - 125 (Juan's number of cards is equal to twice the number of Ryan's cards minus 125) Because that second equation is already in terms of x we can substitute it right into the first equation and solve for y, going through the same process that I described for the first problem.

👍 100 | 👎 -2

The first problem, try to get rid of the y's because one is positive and one is negative. multiply the entire second equation by 2: 2x-2y=-8 Then when you use linear combinations, you get: 5x=-8. x = -8/5. Plug this value back in and you get -8/5 - y = -4. 8/5 +y =4. y = 12/5 Let R = # of baseball cards Ryan has and let J = # of baseball cards Juan has. R+J=880 2R-125=J of 2R-J=+125 You could use linear combinations here and get 3R=1005, R=335 Plug this back in and you get 335+J=880, J=545

👍 96 | 👎 -9

Yep, that's right. And solving for x you get x = -13/2 Each equation is a line, and when you solve for x and y what you're doing is finding the intersection of those lines. Having a y value of 0 just means that the intersection happens to be on the x axis. You can always plug in your x and y values to see if the equation is still true.

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