What will you say to students who do the following in an attempt to help them?
Example 1
1.14 x + 3.42 = 2x - 1
2x - 1.14x - 1 - 3.42 = 0
Example 2
1.338x - 0.699 = 0.778x + 2.334
1.338x + 0.778x = 2.334 - 0.699
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In algebra,like terms can be added or subtracted from one another. the entire equation can then be =o;in quadratic form.solve!. In multiplication and division, the coefficient and the variable are multiplied or divided.In order to simplify algebraic expressions, the distributive law can be used to help us to 'remove the bracket'; once there is one present.
i would advise any person doing an algebraic problem to place all the like terms in one side. or place all the unknown variables in one side and the known in the other then substitute.
in solving this type of algebric equations, it is much easier to place all similar values on one side , paying respect to the changes of the sign(+,-, and x). you then place all common terms on the other side and solve.
eg. 2x + 3 = x + 6
put like terms: 2x-x=6-3
answer = 3
In example 1 you must put all the like terms together so therefore
1.14x+3.42=2x-1 should be 1.14x-2x=-3.41-1
Remember about the POLICE!!! be careful when it comes to changing of positive and negative signs
Example2 is a good example where you have to be careful of changing of positive and negative signs.So therefore to correct this the signs in the equation has to change
1.338x-0.778x=2.334+0.699
in example one all unknown values which is x should go on the left hand side.. den wen changin sides u must remember signs change..
in example 2 the signs were not changed wen the sides were changed so this will cause the whole equation to go wrong
in algebra signs tend to be a funny thing. you have to be careful when moving a term from the LHS to the RHS or vice versa. if the sign infront of the term is positive when u move it to the other side it changes to a negative. also in the examples shown it is wise to have all ur unknowns on one side and ur other values on the other. try not to take any shortcuts, work out one side at a time and you will get ur answer. if u are not sure about ur answer plug it in the original equation for a final check.
lol.....to all the peeps who was so explainin wat the policeman does wham to allya now...:-P
wen you take a term across a = sign the sign always changes!!
so in eg 1 the expression you wud get will be :
1.14x+3.42-2x+1=0
and then you wud put all the like terms lik x on one side of the eq'n
and you wud end up wit:
1.14x-2x=-3.42-1
.....and ten sove for x!
***and then solve for x
lol....my bad :-)
oh wait...in eg.2 the signs heck wrong 2...lol...rem wen you cross the = sign the signs changes
so if there was a - sign infront the term wen you take it over the equal sign it wud change to a + sign!
to get the solution for x it is always easier for me to transpose all of the unknowns to one side of the equal sign and all of the knowns to the other side and then from there just do the addition and subtraction to solve the equation.
But you have to be careful you don't mess up the sign changes. A positive value always changes to a negative one when carried to the next side of the equal sign and vice versa.
The way I have learn about doing algebra is by placing all like terms together and by then simplifying the expression also by heeding the signs(+ and negative as well as the x values.
for instance 4x+12=24-10x
= 4x+10x=24-12
= 14x=12
= 12\14
x = 6/7
in algebra when calculating example1 above you have to rearrange the equation puting all like component on one side and the unlike component on the other side poviding that what was negative is now positive and what was postive turns negative and in the end you just solve.
transpose terms and group x terms
2x-1.14x = 3.42-1
.86x = 2.42
x = 2.42/.86
x = 2.81
When attempting algebraic questions you transpose all the like terms on one side of the equal sign . If you have brackets remove them paying special attention to signs and work from there.
eg. 1.41a-2.35= 1.02a+2.54
Transpose like terms
1.41a-1.02a=2.54+2.35
.39a = 4.89
a = 4.89/.39
a = 12.538
with respect to example 1, u are employing the wrong method. u just making life harder for urself. dont u tink if u had tried to carry all the unknowns on one side and the knowns on the other and solve it wud have been easier??? wat im tryn to say is... 1.14 x + 3.42 = 2x - 1
1.14x - 2x = -1 - 3.42
-0.86x = -4.42
x = 5.14
u understanding wat im sayn???
with respect to example 2, please pay attention to your signs. considering the problem 1.338x - 0.699 = 0.778x + 2.334
remember from my last comment the first ting u do is carry all the unknowns on one side and the knowns on the other and solve. wen u are doin this however, please pay special attention to ur signs. once u are carrying over anyting to the other side of the equation the sign always changes.
therefore
1.338x - 0.699 = 0.778x + 2.334
1.338x-0.778x = 2.334 + 0.699
0.56x = 3.033
x = 5.41
Understand???
Ok when doing these equations the first thing you do is
1) gruop all like terms together
2) put all the numbers with variables on one side of the = and all the numbers without variables on the other side. for example
2x - 1.14x = 3.42 + 1
3) add all the terms on both sides so you will get
0.86x = 4.42
4)then since 3.14x is really 3.14 multiplied by x you can carry across 3.14 so now you will get
x = 4.42/0.86
5)then you just divide and you will get x to be 5.14.
and thats it now you have solved the equation.
yay!!!!!!!!!!!!!!
i agree to all those who spoke about putin all the like variables on one side an being careful about canging signs .... however ... i would jus like to say that wen you solve for x ... to ensure that your ansa is posibly correct .. substitute the ansa you got ..back into the given equation an chec to see if the left side of the = sign corresponds with the right side ... if it doesnt then somtin is wrong an u need to chec over...
rearrange the equation placing like terms on the same side being careful when crossing over the equal sign, since the sign of the value changes. Calculate the x value and the known value on the other side. to calculate the value of x carry over the value multiplied to the other side of the equal
i would say that the reason why we use algebraic expression is to often find or solve for an unknown somethimes more than one unknown. to do this say in the first example u have to take all the common componants one one side and therefore solve for the unknown
i agree with poison.E.g. 1- don't confuse things. once there are no x squared or cubed... terms then leave the variables on one side and the values on the other..
for example 2.
i would tell the person that whenever you carry a variable or term across an equal sign, the sign governing the term changes
i.e
1.338x - 0.699 = 0.778x + 2.334
should be read
1.338x - 0.778x = 2.334 + 0.699
In order to solve example one certain core concepts must be established. Firstly 1.14x means that the constant 1.14 is multiplied by the variable x. this concept must also be applied to 2x in that the constant 2 is multiplied by the variable x. Secondly if a number or variable is 'carried' across the equal sign the mathematical operation it is governed by is reversed. eg. a + would be turned into a - and the other way around and a (/) would turn into a x and the other way around. Thirdly when subtracting or adding components with variables integrated into the component only like variables can work in this scenario. eg.
let a = apples and g = grapes
take this expression 3a - 4g
This expression would translate 3 apples minus 4 grapes which cannot be solved. Lastly it is always good practice to get variables by themselves (on one side of the equation) and constants by themselves. Using thes concepts the question can now be guaged.
first u have to identify any commom terms like the following examples u then get all the like terms on one side of the equation. after do that u add the like terms and then transpose and find x.
i agree with how wong explained the signs changes. it is important to keep like variables together and solve
Put like terms together and substitute.PAY SPECIAL ATTENTION TO YOUR SIGNS PLEASE!
HEY EVERYONE!
Yes in algebra u put all the like terms on one side and the unlike terms on the other side, you not only have to do that you always have to look at your signsif for example you have
2x-4= 6
you keep the unknown on one side and the number on the other side
2x= 6+4
remember to change the sign when carrying a number across to the other side and solve
2x=10
x=10/2
x= 5
PEOPLE REMEMBER THIS RULE ONLY APPLY FOR ADDITION AND SUBTRACTION OF ALGEBRA
WHAT ABOUT MULTIPLICATION AND DIVISION?
Would it be right to group like terms in these questions before trying to solve it?
practice for everyone.. solve..
-2x - 3 + 6x = -2(-x-2+3x)
solve for x
I would tell the the student to take it easy, slow down and take the question in stride. They shouldn't just assume it is a quadratic equation and begin to solve it like that, first see if the situation calls for it.Then begin to solve. first by putting the whole numbers on one side,and the tearms of x on the other. your approach should be thought out carefully before starting any question.
In responce to hellion. Yeah i do think that is the right approach. try the simplest thing first. Right?
The most common of errors are usually the most simple, so simple that they are dismissed or overlooked. One of the main culprits being, the changeing of a sign when being transposed. Just takeing it easy while working and being ever mindfull of these errors will help, and it couldn't hurt to check over a finnished question, would it?
i agree with crimson crock and i don't think that i would hurt to chech over your finished questions well thats for me atlest because it is of your benefit
i think that all like terms should be placed on one side of the equation noting that the signs change wen moved across.in addition the unknowns should be placed usually on the left then it can be solved ex
5x+12 = 4x -15
5x- 4x = -15 - 12
x = - 27
i would advise anybody doing an algebraic problem to place all the like terms in one side or place all the unknowns on one side and the knowns in the otherside then substitute to work out the question
firstly everyone, how can u carry across a term from the LHS to the RHS? In order to move from the LHS to the RHS u have to get rid of the term. So in eg.1 to get rid of 3.42 from the LHS u have to minus 3.42 from the entire equation resulting in -3.42 on the RHS. I think that moose's method is the most simplified in solving the equation.
Marz...
as i have just said if u need to get rid of a term on the LHS side, all u have to do is use the opposite operation in the equation to get it on the RHS.
In example one it is much easier to solve the expression by bringing all the variables on side an all the unknowns on the other side. And then solve for x.
Example 2 is a good way of solving the expression.But you must be careful of changing your signs when putting all the variables on one side and also for the other side.
when u see x terms and the normal terms always put the same terms which are same on one side and the next set on the other side , from there solve respectively
also when the total x term equals to the other side terms, in the equation , divid by the number of the x term [(eg)- 1.338x, divid by 1.338] divid thoughout to get the answer for x
I dont think that example two is correct. I think there is an error when they were bring all the like terms on one side, the sign wasn't change. These are errors that are crusial and should not be meade.
i totally agree with sleepy and alot of people. u have to put all like terms together and then solve its that easy. but don't forget about bodmass at this level of school please!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
its really difficult for me to ensure dat u fully understand algebra via dis blog. u need to go back to ur primary school text book cuz its dere algebra and maths all start. try understanding the basic laws of algebra like the
Commutative law of addition:
a + b = b + a
Associative law of addition:
a + ( b + c )= (a + b) + c
commutative law of multiplication: ab = ba
Associative law of multiplication:
a(bc) = (ab)c
Distributive law:
a(b + c) = ab + ac
MAYBE DAT CUD HELP!!!!!
i think sparkle does not give original ideas and concepts and really just sponges off the ideas of the other bloggers. you think!
In example one(1)I would say that it is not totally wrong because they have grouped their like terms which is a good approach but the common terms can be subtracted to give single terms.Therefore the solution of the equation is not completed.
The second example is wrong however because although the terms were grouped, when the 0.778x and -0.699 is brought on the other sides of the equation the sign changes thus giving -0.778x and +0.699.
Only then could the equation be solved.
I must stand corrected by ichigo's first comment. Actually I now understand that all common terms including those with common letters should be brought one one side and the number terms on the other side of the equation. Only then could the unknown be solved.
cokebaby i tried to do your question. is it correct?
-2x - 3 + 6x = -2(-x-2+3x)
-2x - 3 + 6x = 2x + 4 - 6x
-2x + 6x - 2x + 6x = 4 + 3
8x = 7
x = 7/8
in the examples i think the problem was that no attention was paid to the signs. when moving variables and constants to the other side of the equation you must be very careful that the correct signs are used.when you move a positive number across it becomes negative and when you move a negative number across it becomes positive.you should group all like terms together and then solve this will make it simple and less errors would be made just as was mentioned b4.
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