Algebra Problems

[Green Moss]

Could someone please solve these algebra equations & give a brief explanation:
1) 5(m-2)-4(m+3)=0
2) 5(k+4)-3(k-6)=0
3) 4(y+7)-3(y+5)=0
4) 3(2x-4)-2(4x+5)=0
5) 4x+7=3x-7
6) y=4y+3
7) r=8+9r
8) w=5q-R
9) p=4m+8n
10) w+7=3w-1

My son is stuck with half of his homework & any help would be appreciated........Thanks

[mutt53]

1) m =22

[syruphead]

10.
w = 4

[syruphead]

7.
r = -1

[syruphead]

6. y= -1

[syruphead]

5.
x= -14

[johnny.5]

6 . y= -1
-1x4=-4
-4+3=-1

[Fibonacci]

The first few questions require the brackets to be expanded (multiplied out) and then 'like' terms collected before solving the equations
1) 5m - 10 - 4m - +12 = 0 so 1m - 22 = 0 , m =22
2) 5k + 20 - 3k - -18 = 0 so 2k + 38 = 0 , k = -19
3) 4y + 28 - 3y - +15 = 0 so 1y + 13 = 0, k = -13
4) 6x - 12 - 8x - +10 = 0 so -2x - 22 = 0, x = -11
The next few require the same operation to be carried out to each side of the equation as it is solved.
5) 4x+7=3x-7
(-3x) (-3x)
x + 7 = -7
(-7) (-7)
x = -14

6) y=4y+3 can be rewritten
4y + 3 = y
(-y) ( -y)
3y - 3 = 0
(+3) (=3)
4y = 3 , so y = 1

7) r=8+9r can be rewritten
9r + 8 = r
(-r) (-r)
8r + 8 = 0
(-8) (-8)
8r = -8, so r = -1

8) w=5q-R
9) p=4m+8n
cannot be solved as they srae stated as there is more than one unknown quantity in each question.

10) w+7=3w-1 can be rewritten
3w - 1 = w + 7
(-w) (-w)
2w - 1 = 7
(+1) (+1)
2w = 8, so w = 4

[syruphead]

3.
y= -13

[syruphead]

well done fibonacci, youv'e got the right name for this sort of thing - just a theory I have !

[Buenchico]

(2-part post):

Questions 8 and 9 don't have numerical answers and, as they stand, don't make any sense at all. (Question 8, for example could state "Rearrange the equation to make R the subject of the formula" but, without such a question or further information about the numeric values of two of the lettered terms, it's currently 'impossible').

Answers to the others (with explanations):

Q1. First get rid of the brackets by 'expanding' the terms. (i.e. multiply everything inside each bracket by what's outside of it). We then get:

5m - 10 - 4m -12 = 0

Now group the terms, thus:

m - 22 = 0

Add 22 to both sides (yo leave m on its own):

m = 22


Q2: Multiply out (as before):
5k + 20 -3k + 18 = 0
Group the terms:
2k + 38 = 0
Take 38 from both sides (to leave 2k on its own):
2k = -38
Divide everything by 2 (to leave k on its own):
k = -19

[Buenchico]

Q3: Multiply out:
4y + 28 - 3y -15 = 0
Group the terms:
y + 13 = 0
Subtract 13 (to leave y on its own):
y = -13

Q4: Multiply out:
6x - 12 - 8x -10 = 0
Group the terms:
-2x -22 = 0
Add 22 to both sides (to leave -2x on its own):
-2x = 22
Divide by -2 (to leave x on its own):
x = -11

Q5: Subtract 3x from both sides (so that x only appears once in the equation):
x + 7 = -7
Subtract 7 from both sides (to leave x on its own):
x = -14

Q6: Subtract 4y from both sides (so that y only occurs once in the equation):
-3y = 3
Divide by -3 (to leave y on its own):
y = -1

Q7: Subtract 9r from both sides (so that r will only occur once in the equation):
-8r = 8
Divide by -8 (to leave r on its own):
r = -1

Q10: Subtract 3w from both sides (so that w only occurs once in the equation):
-2w + 7 = -1
Subtract 7 from both sides (to leave -2w on its own):
-2w = -8
Divide by -2 (to leave w on its own):
w = 4

Chris (a former maths teacher!)

[johnny.5]

I hate algebra

Am I right ?
I'm not wrong !

[Green Moss]

Thank you all very much for your answers, especially Fibonacci & Buenchico (as usual) - much appreciated, plus, Buenchico all your help on other topics. I bet you were a great teacher!