1.
3x+ 5 = 197 subtract 5 from both sides
3x + 192 divide both sides by 3
x = 64
2.
1/3(1/2x) - 11 = 38 simply
1/6x - 11 = 38 add 11 to both sides
1/6x = 49 multiply both sides by 6
x = 294
3.
3/4x - 8 = 91 add 8 to both sides
3/4x = 99 multiply both sides by 4/3
x = 132
4.
x + (x + 2) + (x+4) = 105 simply
3x+6=105 subtract 6 from both sides
3x=99 divide both sides by 3
x = 33
so 33, 35, 37
33+35+37 = 105
5.
x + (x + 1) + (x+2) + (x + 3) + (x + 4) = 195 simply
5x+10=195 subtract 10 from both sides
5x =185 divide both sides by 5
x = 37
so 37, 38, 39, 40, 41
37 + 38 + 39 + 40 + 41 = 195
Hope it helps
2006-12-05 07:32:23
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answer #1
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answered by Richard 7
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1) five more than three times a number is 197.Find the number.
Answer: You substitute x for the unknown number and create an equation. So the equation becomes 3x+5=197 . After creating the equation you solve for x. First you want to get x alone so you subtract five from both sides of the equal side. Then you must divide 192 by 3 to get alone x and find x. When you divide you get 64 which is your answer. To check the answer plug 64 into the equation you created. 3(64)+5=197
3x+5=197
-5= -5
3x=192
x=64
2)x=98 but I am not sure if the answer is correct.
3) Again I am not positive this in correct but I believe the answer may be. x=74.5
4) For numbers four and five I just used the guess and check method and found that 33,35 and 37=105
5)35,37,39,41,43=195
2006-12-05 15:28:10
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answer #2
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answered by cdc5112 1
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formula to solve #1: 5 + 3(x) = 197
197 - 5 = 3(x)
192 divided by 3 = x = 64
formula to solve # 2: (x/2) divided by 3 -11 = 91
(x/2) divided by 3 = 102
x/2 = 102 times 3 = 306
x = 306 times 2 = 612
formula to solve # 3: 3x/4 - 8 = 91
91 + 8 = 3x/4
99 =3x/4
99 times 4 = 3x
396 divided by 3 = x = 132
Number 4: 33, 35, 37
Number 5: 35, 37, 39, 41, 43
I don't remember exactly how to get integers, what I did was divide the resulting number by how many integers you needed, that way you get the middle one, then just figure out the consecutive ones before and after.
Hope this helped, and try hitting the books harder from now on.
2006-12-05 15:19:32
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answer #3
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answered by guicho79 4
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1.) five more (+5) than 3 times a number (3x) is 197(=197)
3x+5=197 you can figure it from there.
2. Half of x (1/2x)divided by 3 minus 11(- 11) is 38(=38).
(1/2x divided by3) - 11 = 38
3. Eight less (-8) than three quarters of a number (3/4x) is 91(=91)
3/4x - 8 = 91
There's the first three to get you started.
2006-12-05 15:10:42
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answer #4
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answered by indigo302 2
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1.)five more than three times a number is 197.Find the number.
192 x X = +5 64!
2.)When one third of half a number is decreased by 11,the result is 38.find the number. [PDF] Applications of Linear EquationsFile Format: PDF/Adobe Acrobat - View as HTML
(3) If a number is decreased by 5, the result is twice the original number. ... (7) If the width of a rectangle is 5 cm more than one-half its length and ...
www.math.wfu.edu/Math105/Applications%20of%20Linear%20Equations.pdf - Similar pages
3.)Eight less than three quarters of a number is 91.find the number.
4.)Find three consecutive odd integers whose sum is 105.
5.)Find five consecutive integers whose sum is 195.
2006-12-05 14:56:29
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answer #5
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answered by god knows and sees else Yahoo 6
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1)3x+5=197
3x=192
x=64
2)(x/2)/3 -11=38
x/6 -11=38
x/6=49
x=294
3)
3x/4 -8 =91
3x/4 = 99
3x=396
x=132
4) x + (x+2) + (x+4) = 105
3x +6 =105
3x=99
x=33
33 +35 +37 = 105
5) x + (x+1) +(x+2) + (x+3) +(x+4) = 195
5x +10 = 195
5x=185
x= 37
2006-12-05 15:21:26
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answer #6
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answered by Alex 2
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1) 3x + 5 = 197
2) (1/3 (x/2)) - 11 = 38
3) 3/4x - 8 = 91
That should get you started. Solve for x, or someone else can take it from here.
Can't help you with the last two. Sorry.
2006-12-05 14:56:43
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answer #7
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answered by hey u 3
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1) 3x+5=197
3x=192
x=64
2) x/6-11=38
x/6 = 49
x=294
3)3x/4-8=91
3x/4=99
x=132
4) 3x = 105
x=33,35,37
5)5x=195
x=37,38,39,40,41
2006-12-05 14:58:07
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answer #8
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answered by Paul B 3
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1)X=64
2)X=147
3)122
2006-12-05 14:58:41
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answer #9
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answered by Alex F 3
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