SOLUTION: Solve the system by addition. 5x – 3y = 5 x – 5y = 1 Solve the system by addition. 3x + 8y = 7 6x – 4y = –1

Algebra ->  Linear-equations -> SOLUTION: Solve the system by addition. 5x – 3y = 5 x – 5y = 1 Solve the system by addition. 3x + 8y = 7 6x – 4y = –1       Log On


   

Question 89301: Solve the system by addition. 5x – 3y = 5 x – 5y = 1

Solve the system by addition. 3x + 8y = 7 6x – 4y = –1
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations

5%2Ax-3%2Ay=5
1%2Ax-5%2Ay=1

In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for y, we would have to eliminate x (or vice versa).

So lets eliminate x. In order to do that, we need to have both x coefficients that are equal but have opposite signs (for instance 2 and -2 are equal but have opposite signs). This way they will add to zero.

So to make the x coefficients equal but opposite, we need to multiply both x coefficients by some number to get them to an equal number. So if we wanted to get 5 and 1 to some equal number, we could try to get them to the LCM.

Since the LCM of 5 and 1 is 5, we need to multiply both sides of the top equation by 1 and multiply both sides of the bottom equation by -5 like this:

1%2A%285%2Ax-3%2Ay%29=%285%29%2A1 Multiply the top equation (both sides) by 1
-5%2A%281%2Ax-5%2Ay%29=%281%29%2A-5 Multiply the bottom equation (both sides) by -5


So after multiplying we get this:
5%2Ax-3%2Ay=5
-5%2Ax%2B25%2Ay=-5

Notice how 5 and -5 add to zero (ie 5%2B-5=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%285%2Ax-5%2Ax%29-3%2Ay%2B25%2Ay%29=5-5

%285-5%29%2Ax-3%2B25%29y=5-5

cross%285%2B-5%29%2Ax%2B%28-3%2B25%29%2Ay=5-5 Notice the x coefficients add to zero and cancel out. This means we've eliminated x altogether.



So after adding and canceling out the x terms we're left with:

22%2Ay=0

y=0%2F22 Divide both sides by 22 to solve for y



y=0 Reduce


Now plug this answer into the top equation 5%2Ax-3%2Ay=5 to solve for x

5%2Ax-3%280%29=5 Plug in y=0


5%2Ax%2B0=5 Multiply



5%2Ax=5-0 Subtract 0 from both sides

5%2Ax=5 Combine the terms on the right side

cross%28%281%2F5%29%285%29%29%2Ax=%285%29%281%2F5%29 Multiply both sides by 1%2F5. This will cancel out 5 on the left side.


x=1 Multiply the terms on the right side


So our answer is

x=1, y=0

which also looks like

(1, 0)

Notice if we graph the equations (if you need help with graphing, check out this solver)

5%2Ax-3%2Ay=5
1%2Ax-5%2Ay=1

we get



graph of 5%2Ax-3%2Ay=5 (red) 1%2Ax-5%2Ay=1 (green) (hint: you may have to solve for y to graph these) and the intersection of the lines (blue circle).


and we can see that the two equations intersect at (1,0). This verifies our answer.



----------------------------------------------------

Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition


Lets start with the given system of linear equations

3%2Ax%2B8%2Ay=7
6%2Ax-4%2Ay=-1

In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for y, we would have to eliminate x (or vice versa).

So lets eliminate x. In order to do that, we need to have both x coefficients that are equal but have opposite signs (for instance 2 and -2 are equal but have opposite signs). This way they will add to zero.

So to make the x coefficients equal but opposite, we need to multiply both x coefficients by some number to get them to an equal number. So if we wanted to get 3 and 6 to some equal number, we could try to get them to the LCM.

Since the LCM of 3 and 6 is 6, we need to multiply both sides of the top equation by 2 and multiply both sides of the bottom equation by -1 like this:

2%2A%283%2Ax%2B8%2Ay%29=%287%29%2A2 Multiply the top equation (both sides) by 2
-1%2A%286%2Ax-4%2Ay%29=%28-1%29%2A-1 Multiply the bottom equation (both sides) by -1


So after multiplying we get this:
6%2Ax%2B16%2Ay=14
-6%2Ax%2B4%2Ay=1

Notice how 6 and -6 add to zero (ie 6%2B-6=0)


Now add the equations together. In order to add 2 equations, group like terms and combine them
%286%2Ax-6%2Ax%29%2B%2816%2Ay%2B4%2Ay%29=14%2B1

%286-6%29%2Ax%2B%2816%2B4%29y=14%2B1

cross%286%2B-6%29%2Ax%2B%2816%2B4%29%2Ay=14%2B1 Notice the x coefficients add to zero and cancel out. This means we've eliminated x altogether.



So after adding and canceling out the x terms we're left with:

20%2Ay=15

y=15%2F20 Divide both sides by 20 to solve for y



y=3%2F4 Reduce


Now plug this answer into the top equation 3%2Ax%2B8%2Ay=7 to solve for x

3%2Ax%2B8%283%2F4%29=7 Plug in y=3%2F4


3%2Ax%2B24%2F4=7 Multiply



3%2Ax%2B6=7 Reduce



3%2Ax=7-6 Subtract 6 from both sides

3%2Ax=1 Combine the terms on the right side

cross%28%281%2F3%29%283%29%29%2Ax=%281%29%281%2F3%29 Multiply both sides by 1%2F3. This will cancel out 3 on the left side.


x=1%2F3 Multiply the terms on the right side


So our answer is

x=1%2F3, y=3%2F4

which also looks like

(1%2F3, 3%2F4)

Notice if we graph the equations (if you need help with graphing, check out this solver)

3%2Ax%2B8%2Ay=7
6%2Ax-4%2Ay=-1

we get



graph of 3%2Ax%2B8%2Ay=7 (red) 6%2Ax-4%2Ay=-1 (green) (hint: you may have to solve for y to graph these) and the intersection of the lines (blue circle).


and we can see that the two equations intersect at (1%2F3,3%2F4). This verifies our answer.