Solving systems of equations algebraically?

Question

My new math teacher this year is horrible at explaining things and i learned how to solve equations algebraically last year and now she just toally confused me. If someone can tell me step by step what to do for example with this equation


3x+y=16
2x+y=11

Or this one i'm confused by..

x+2y=8
x-2y=4


If someone can just tell me what your supposed to do step by step so i understand that would be great.

Answer 1

You could solve by substitution, but I think that solving by elimination is easier for these systems.

The idea is to get one of the variables to cancel out.

3x+y=16
2x+y=11 subtract this from the first eqn
------------
x=5 substitute this back in and solve for y

3(5)+y=16
y=1

So, the solution is
x=5 and y=1

x+2y=8
x-2y=4 add them together
----------
2x=12
x=6 substitute x and solve for y

6+2y=8
2y=2
y=1

So, the solution is
x=6 and y=1



You can see that everyone has different answers. To be sure that your answer is correct, you should plug your numbers back into both of the original equations to verify your answer.


3(5)+1=16 true
2(5)+1=11 true

6+2(1)=8 true
6-2(1)=4 true

Answer 2

Sorry about your math teacher! So what you do on the first one is....you're either using elimination or substitution I'm guessing by the problem. In this case you have two variables that are ones so substitution would be easier. So on the first (or second. Whatever you choose) Put y on one side of the = sign and the other stuff on the other side. So ... y= -3x +16. I just moved the 3x over. Then you substitute -3x+16 into the other equation for y. 2x + (-3x+16) = 11. Now solve normally. So... (2x + -3x is -x by the way) -x +16 = 11. Subtract and you get -x= -5. Change that to x=5. It's the same thing. Now all you need is the y. So plug the 5 into one of the two equations and solve again. I'm going to use the second one. 2(5) + y = 11. So y= 1. Good luck and I hope I helped!

Answer 3

There are a few different options that you can use for solving systems of two equations in two variables. The two most commonly used are addition and substitution.

If I were to solve the first set by addition:
3x+y=16
2x+y=11
(This is a really good set of equations to solve using the addition method since both equations have a y variable with a coefficient of 1)

If I were to multiply equation 1 in this set by -1
-3x-y=-16
and keep equation 2 the same
2x+y=11
I now have as my system:
-3x-y=-16
2x+y=11
add the two equations together, term by term:
-3x+2x-y+y=16+11
now simplify the like terms:
-x=27 therefore x=-27
Since we know by the original equations that 2x+y = 11 we also know that 2(-27)+y=11
Solve for y and we get -54+y=11; y=65
solution x=-27; y=65 sometimes written as an ordered pair (-27, 65)

Set two is also an excellent one to solve by addition since those "2y" terms have opposite signs (big hint!!!!!!). However, another approach commonly used is substitution.
In substitution, you solve for one variable in one equation (in terms of the other)
(e.g. x+2y=8 becomes x=8-2y)
Now, we can substitute 8-2y for x in the second equation:
8-2y-2y=4; combine like terms to get 8-4y=4
-4y=-4 so y=1
Substitute 1 for y back in the first equation
x+2=8 so x=6
solution (6,1)

Hope this helps.

Answer 4

For both of these, the easiest method is elimination:

subtract the first equation from the second:
3x+y = 16
-(2x +y = 11)

3x+y=16
-2x-y=-11

x=5
then plug into either equation to get the y:
2(5) +y =11
y=1



x+2y=8
-(x-2y = 4)

x+2y=8
-x+2y=-4

4y=4
y=1
then plug into either equation to get the x:
x+ 2(1) =8
x=6

Answer 5

Remember that the basic rule of algebra is that you can apply an equivalent operation to both sides of an equation. So for the first problem, you might try subtracting the same quantity (or two equivalent quantities) from both sides of the top equation. You can subtract 2x+y from 3x+y. Instead of subtracting 2x+y from 16, use its equivalent value of 11. So you are left with x=5.

When you get to more complicated equations, you'll want to solve one equation for one variable, then substitute into the second equation. For example, y=16-3x, then 2x + (16-3x) = 11, then -x+16=11, then -x=-5, then x=5.

Answer 6

2. x+2y & x - 2y = 4 > Solve for "x" or "y" in either equation. We'll solve for "x" in the 2nd equation.

First: add "2y" with both sides (when you move a term to the opposite side, always use the opposite sign).

x - 2y+2y = 4+2y
x = 4+2y

SEc: substitute "4+2y" with "x" in the 1st equation.

4+2y+2y = 8
4+4y = 8 (subtract 4 from both sides).
4 - 4+4y = 8 - 4
4y = 8 - 4
4y = 4 (divide both sides by 4).
4y/4 = 4/4
y = 4/4 = 1

Third: substitute 1 with "y" in the 2nd equation.

x - 2(1) = 4
x - 2 = 4 (add 2 to both sides).
x - 2+2 = 4+2
x = 6

Solution: (1, 6) > P.S. Follow the method to solve # 1 :-)

Answer 7

Question 1
3x + y = 16
-2x - y = - 11----ADD
x = 5

15 + y = 16
y = 1

x = 5 , y = 1

Question 2
x + 2y = 8
x - 2y = 4-----ADD
2x = 12
x = 6

6 + 2y = 8
2y = 2
y = 1

x = 6 , y = 1

Answer 8

5x + 2y = 15 -5x - 2y = -5 _____________ 0 = 10 you remedy utilising combining words, you upload the two equations. 0=10 skill this technique of equation has no answer in any respect, through fact the slopes of the two lines are an identical, in different words are parallel lines, so there is not any longer factor of interception

Answer 9

3x+y=16
2x+y=11

if you multiply either equation by -1 the ys will cancel...

3x + y = 16
-1(2x + y = 11)

3x + y = 16
-2x - y = -11

x = 5

now solve for y
3x+y=16
3(5)+y=16
15 + y = 16
-15 -15
------------------
y = 1

so x = 5 and y = 1









x+2y=8
x-2y=4

if you notice 2y and -2y cancel... combine the remaining equestions...

2x = 12

get x by itself by dividing by 2

2x = 12
--- ---
2 2
x = 6

Now solve for y using x = 6 choose either original eqution....

x + 2y = 8
(6) + 2y = 8
-6 -6
-------------------
2y = 2
--- ---
2 2
y = 1

So x = 6 and y = 1 for the second problem

Answer 10

subtract the first and second equation
x = 5
y = 16 - 15 = 1

2x = 12
x = 6
y = (4 - 6) /2 = -1