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2X+5Y=412X+Y=13solve using combination method

if the system has an infinite solution set, specify it and give three solutions


Where does 412x separate?  Is it 4, 12x or 41, 2x or 412, x?


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3 Answers

2x + 5y = 41       

-(2x + y = 13)

2x + 5y = 41

-2x - y =- 13


4    4


2x + 5((7) = 41




2x + 5y = 41     Equation 1

2x + y = 13       Equation 2

The combination method combines the two equations similar to the elimination method.

2x + 5y = 41  (In this case, it will be easier to leave the first equation alone.)

-2x - y = -13   (Multiply your second equation by a -1 to make one of the variables equal but opposite, in this case, x.  All we are doing is changing the signs of the second equation.

Now we combine the two equations.  We write it out like this:

2x + 5y + (-2x - y) = 41 + (-13)

We now need to combine like terms and find that when we multiplied the first equation by 6, the x variables will cancel.  Don't forget to watch your negative signs.

2x - 2x + 5y - y = 41 - 13      Combine the like terms. 

0x + 4y = 28     We don't need to keep the 0x in the equation anymore.  Now we divide each side by a 4 to isolate the y variable.

We find that y = 7.  Since we have one variable, we can now plug this value in to either of the two original equations.  I will choose equation 2 because it's a little bit easier, but both will work.

2x + (7) = 13   Plug in 7 wherever you see a y variable.

2x = 6             Subtract 7 from both sides.

x = 3               We find our answer to be 3.

Now we have both our x value and our y value.  Our answer is x=3, y=7.

rearrange the second to be y=13-2x

then substitute this back into the first for y

thus, 2x+5(13-2x)=41

which reduces to 2x+65-10x=41 further 65-8x=41 then solving for x, we get x=3

plug x=3 back into both

the first one is then 5y=35 which tells us y = 7

the second is then 6+y=13 which also reduces to y=7

so there is only one unique solution to this system and it is x=3, y=7

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