solve the equations for X

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Hi I am Andrew's mom. He is 14 and we live in the Midwest. It snowed last night for the first time this season. It’s very pretty outside.

We are trying to help Andrew, during the past few months the school has been working with him on beginning algebra and we need some guidance. We thought this might be a good site to seek assistance.

Andrew just started pre-Algebra this year and is doing very well. However solving x + 5 = 27 equations are difficult. He can cross multiply and divide to solve the proportions, simplify expressions by combining like terms, change fractions to decimals, etc.

Any suggestions for solving for X + 5=27
X-3=10 type problems?


I was thinking about a step-by-step example.
P.S. Don’t’ know if this would help, but for addition and subtraction problems Andrew uses a calculator. Thank you. Any assistance would be appreciated.

The goal is to get the letter on one side, the numbers on the other, in the most basic of terms.

X + 5 = 27
- 5 - 5
--------------------
X = 22


X - 3 = 10
+ 3 = + 3
--------------------
X = 13

As the equations get more complex, there are more rules and steps to follow... but I'm not sure what more you are looking for. You might try doing example questions with all the numbers there, but a cup or shell covering what is supposed to be the 'variable'. By moving the pieces, he can then lift the shell and 'see' if he got the right answer.


M.

Just remember that whatever you do to one side of the equation, you can do to the other. For example:

x=10
x+3=10+3
x+3=13
2(x+3)=2(13)
2x+6=26
2x+6-6=26-6
2x=20
2x/2=20/2
x=10

As you can see, just make sure that whatever you do to one side of the equation, you do to the other.

[quote="MomofAndrew"]Hi I am Andrew's mom. He is 14 and we live in the Midwest. It snowed last night for the first time this season. It’s very pretty outside.

We are trying to help Andrew, during the past few months the school has been working with him on beginning algebra and we need some guidance. We thought this might be a good site to seek assistance.

Andrew just started pre-Algebra this year and is doing very well. However solving x + 5 = 27 equations are difficult. He can cross multiply and divide to solve the proportions, simplify expressions by combining like terms, change fractions to decimals, etc.

Any suggestions for solving for X + 5=27
X-3=10 type problems?


Sure!

For x+5=27
Step one: Subtract 5 from both sides to isolate x.
x+5-5=27-22

Step 2: Solve for x
x=5

If x is added to a number, whatever side x is on, you subtract that amount from both sides of the equation. If x is subtracted from a number, you add that number to both sides of the equation

slightly different approach:

get rid of anything that is on the x side by employing opposite operation

for example:

X + 5=27

on left side we have +5 so it transferes to the right side and becomes -5 thus we have x=27-5=22

or for X-3=10

-3 from left side goes to the right side and becomes +3 so x=10+3=13

it is slightly more complicated with * and / operations but if careful should work.

Bit off there.

X + 5 = 27
X + 5 - 5 = 27 - 5
X + 0 = 22
X = 22
22 + 5 = 27
27 = 27 = true

If he's still having trouble, try applying (if possible) to an interest.

For example, if his favorite team has 27 points in a game, and the other team has 5 points fewer (or 5 points down, assuming the lingo), how many points does that team have?
or . . . his team is playing the game and has 27 points, and if the other team had 5 more points, they would be tied. how many points does that team have? I usually see math as pattern and logic thinking; if this problem use this method; with this setup of numbers, you get this answer, etc. Have fun with it as its not nearly as hard as some think (unless there is some dx i dont know about ><). Good luck.

I learned to solve equations in a slightly different way. Instead of doing the same operation to both sides, I was taught to "move the number with the opposite sign to the opposite side of the equation". So,

x + 4 = 10

becomes

x = 10 - 4
and then
x = 6

Check the equation by substituting the solution for X.
6 + 4 = 10
10 = 10 is TRUE

because when 4 moves from one side of the equation to the other, it changes its sign. The add/subtract sign before the number belongs to that number and makes it positive or negative. The end result is the same as performing the same operation on both sides, but in this case, the process of arriving at it is different.

Another example,
x - 7 = 13
x = 13 + 7
x = 20
Check the equation,
20 - 7 = 13
13 = 13 is TRUE

For multiplication and division, the traditional "perform the same operation on both sides" method is used to to cancel out the multiplier and arrive at X.
So,

3x + 3 = 18
3x = 18 - 3
3x = 15
(3x)/3 = 15/3
x = 5
Check it,
3 * 5 + 3 = 18
15 + 3 = 18
18 = 18 is TRUE

So, here's the gist of how to "move the number with the opposite sign to the opposite side of the equation". I found this method much more clear and understandable than the traditional American method. My middle school math teacher was from Poland, and she showed the class how to do this; apparently, that's how they teach algebra back there, and maybe other parts of Europe too. She let the students use either one of the methods she taught, because she could understand both.

Perhaps you can show your son this nontraditional method (for America, at least) of doing equations. Perhaps he finds adding and subtracting the same numbers on both sides too cumbersome or what have you. The European method seems a lot more streamlined, in my opinion, and that's how equations work in my mind. Maybe your son will agree.