# How do I solve this Algebra problem?

The lesson that I am learning is: Solve Linear Systems by Adding or Subtracting and I am having some trouble with problem #20

Here it is:

11y-3x=18

-3x= -16y+33

I am getting like a decimal. Any help would be great thanks!

Relevance

reorganize and solve

-3x=-16y+33 becomes 16y -3x = 33

subtract second from first to change sign

11y-3x = 18

16y-3x = 33 subtract

_______________

-5y = -15

y = -15/-5 = 3

substitute back into the equations

-3X = -16(3) + 33

-3X = -48 + 33

-3x = -15

x = 5

check 11(3) - 3(5) = 18

33 -15 = 18

18 = 18

-3(5) = - 16(3) + 33

-15 = -48 + 33 = -15

• Lucy
Lv 7

The answer definitely does not include decimals. Here is how you do it:

Given:

11y - 3x = 18

-3x = -16y + 33

First, write both of these equations so that the variables line up with each other.

-3x + 11y = 18

-3x + 16y = 33

Multiply the first one by -1.

-1(-3x + 11y) = -1(18)

-1(-3x + 11y) = -18

Distribute.

-1(-3x) - 1(11y) = -18

3x - 11y = -18

3x - 11y = -18

-3x + 16y = 33

3x - 11y = -18

-3x + 16y = 33

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

5y = 15

Divide both sides by 5.

5y / 5 = 15 / 5

y = 3

Plug this back into one of the equations and solve for x.

-3x = -16y + 33

-3x = -16(3) + 33

-3x = -48 + 33

-3x = -15

-3x / -3 = -15 / -3

x = 5

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

CHECK:

You can always check your answer by plugging in the values you found back into the original equations to find out if both sides equal each other.

11y - 3x = 18

11(3) - 3(5) = 18?

33 - 15 = 18?

18 = 18?

true

-3x = -16y + 33

-3(5) = -16(3) + 33?

-15 = -48 + 33?

-15 = -15?

true

Source(s): For more help on elimination: http://www.purplemath.com/modules/systlin5.htm
• Anonymous

actually you could use the addition by subtraction method

11y-3x=18

-3x= -16y+33

-3x+11y=18

-3x +16y=-33

+ -3x+11y=18

(- ) 3x -16y=-33

-5y=-15

-5 -5

y= 3

then substitute to get the value of x

-3x= -16y+33

-3x= -16(3)+33

-3x= -48+33

-3x=-15

-3 -3

x=5

i would make additional information later i still have my class.. hang on there.. God bless

Well the equation for a slope is y=mx+b. so first off get it to this point by rearranging the equation. b is the plain old number. nothing special about it. the mx part is the x and the m which is the coefficient. so for example 5x would = mx because the 5 is the coefficient. and y is...y. So for the first one you add 3x to both sides and you get 11y=3x+18. Hope I could help.

heres how u solve. put all numbers and variables aligned like this:

16y-3x=33

11y-3x=33

make sure u cancel one of them out. multiply one of them by -1.

16y-3x=33

-11y+3x=-33

16y-11y=5y. x's cancel and so do the regular numbers.

5y=0

plug y in.

16(0)-3x=33

-3x=33

x=11

• Anonymous

-3x + 11y = 18

3x - 16y = -33

-5y = -15

y = 3

-3x + 11(3) = 18

-3x = 18 - 33

-3x = -15

x = 5

11y-3x=18

or

-3x = -11y+18

-3x= -16y+33

Now

-11y+18 = -16y+33

5y = 15

y = 3

Substitute:

x = 5

11y-3(-16y+33)=18

11y-3x=18..............Eq.1

-3x= -16y+33..........Eq.2

Re-organize both equations

-3x+11y=18............Eq.1

-3x+16y=33............Eq.2

-3x+11y= 18...........Eq.1

3x-16y= -33...........Eq.2 X -1

-5y=-15

5y=15

y=3

Substitute 3 for y in either original equation.

I'll use -3x+11y= 18

-3x+11(3)=18

-3x+33=18

-3x=18-33

-3x=-15

3x=15

x=5

Solution is x=5, y=3

• Anonymous

OK. You first want to solve for a variable in the first equation. Either X or Y. You get everything on one side equal to That variable. Then plug it into the 2nd equation. Whatever was on the other side you will plug into the variable you solved for. There should only be one variable in your equation. Then you solve for that variable. Then plug that into the original first equation.

Source(s): MATHHHH