equations Flashcards
If x³ +b³ = (x+b) (x²-xb + 9) and b>0 , what is the value of b
3
x³ (x² - 10) = -9x
if x>0, what is the solutions
c(ax−b)−d(b−ax)
Which of the following is equivalent to the expression above?
(ax-b)(c+d)
(ax-b)² (c+d)
(b-ax)²(c-d)
(ax-b)(b-ax)(c-d)
numberx^4 - number
This means, when an equation does’nt have a number with x , the factored form is always:
256a^4 - 1
(ax +b) (ax-b) (cx +b) ²
4x + 1) (4x - 1) (16x² +1
(y² - 32y +256 - k) (y² -32y+256 + k) = (y-16)^4 - 100
what is the value of k
(y² - 32y +256 ) = (y-16)²
Therefore, - k + k = 100
s³t³ + 1
a) (st-1) ( s²t²+st + 1)
b) (st- 1 ) (s²t² - st - 1)
c) (st-1) ( (st)²-st + 1)
c) (st+1) ( (st)²-st + 1)
1) Prove one equation and analyze how to get the correct answer
a) (st-1) ( s²t²+st + 1) : s³t³ +s²t² +st - s²t² - st - 1 -> s³t³ -1
“1” need to be +1 -> So, the last 1 should be both positives
c) (st+1) ( (st)²-st + 1)
w= 6+ 3xy/75
The equation above gives the quantity w in terms of the quantities x and y Which of the following equations correctly expresses yin terms of w and x?
a) y= 25w + 150 / x
b) y= 25w - 150 / x
c) ) y= 25w -2 / x
d) a) 25w + 2 / x
w= 6+ 3xy/75
(w - 6 ) 75 = 3xy
(75w - 75.6)/3 = xy
25w - 150) /x = y
y-b / x-a = m a= a) -mx + y - b /m b) mx + y - b /m c) y - b /m d) -y +b /m
- Eliminate the denominator first
y-b = m (x-a) -> y-b = mx - ma - Isolate a
y-b-mx/ m = -a - Multiplies -1 , so a be positive
(y-b+mx/ m) . -1 = a => - y-b+mx/ m
(1+1 /2) . -1 : the -1 multiplies only the fraction as a whole or also affects the sum inside “1+1”
a(1-x) + 5y = 18
MULTIPLIES THE FRACTION AS A WHOLE
18 - 5y = a(1-x)
18-5y/a - 1 = -x
(18-5y/a - 1). -1 = x => - 18-5y/a + 1
T= √(uqx)² + (To)²
Which of the following equations correctly gives x in terms of T, To, u?
a) x = T-To/uq
b) x = T-(To)²/uq
c) x= √T²-(To)²/uq
c) x= √T²-(To)²/uq
T= √(uqx)² + (To)² => T= (uqx)² + (To)²
√T - To² = √(uqx)²
v = 331.3 √1 + T/ 273.15
√ extends to the T fraction
(v / 331.3 - √1 )² = + (√T/ 273.15 )²
(v²/ 331.3² - 1 ) = + T/ 273.15
(v²/ 331.3² - 1 ) 273.15 = + T
(x-h)² + ( y - k)² = r²
a) y = r-x + h + k
b) y = r - (x-h)² + k
c) y = +- √r² - (x-h)² + k
r² - (x-h)² = (y-k)²
+ - √r² - (x-h)² = √(y-k)²
+ - √r² - (x-h)² + k = y
k² = m² + n²
a) m = k - n
b) m = √k-n
b) m = √k²-n²
b) m = √k-n
the square extends to k-n
The area, A, of the unshaded circular ring shown at left is given by the equation A= π(R²-r^²)
where R is the radius of the larger circle, and r is the radius of the smaller circle. Which of the following equations correctly gives R in terms of A and r?
a) R= √A / pi + r²
b) R= √A / pi + r² ( R= √extends to r²
c) R= √A / pi - r² ( R= √extends to r²
A/pi + r² = R²
√A/pi + r² =R B
n = (1.96o / E) ²
a) o = √En/1.96
B) o = E. √n/1.96
C) o = 1.96n /E
nE² /1.96²= o²
Simplify
√nE²/√1.96² = √o² -> √nE/1.96
√10X - 2xy = 4x
√ extends to -2xy
a) y = - 18x
b) y = -8x + 5
c) y = -2x + 5
d) -16x² - 5x
1) Eliminate the ratio
(√10X - 2xy)² = (4x )² -> 10x - 2xy = 16x²
2) Simplify by dividing 2
5x - xy = 8x²
3) Isolate the result
(8x² - 5x ) / - 1 x = -8x² + 5
let’s say we have A and B in the equation. But, there are two of B in the fraction position
R = C-P/ P
1) Eliminate the denominator by multiplying it both sides
RP = C- P
2) Again, eliminate the B ( who has a twin in the equation ) by adding B both sides
RP + P = C
3) Isolate the common : P ( 1+ R) = C
4) Isolate the desired one :P = C / (1+R)
let’s say we have A and B in the equation. But, there are two of B in the fraction position
C= ab / a + b
a =
1) Eliminate the denominator by multiplying both sides
C ( a+b) = ab -> Ca + Cb = ab
2) Eliminate the one that has twin and it’s going to be the result (f(x)
Cb = ab - Ca
3) Isolate the common
Cb = a (b-C)
4) Isolate the result
Cb/ b - C = a
1/f = 1/o + 1/i
o =
1/f - 1/i = 1/o ->
2) MMC
i - f /fi = 1/o
3) Eliminate the numerator 1 by inverting the ratio : potent ^-1
(i - f /fi)^-1= (1/o)^-1 => fi/ i - f = o
V = C ( 1-r)^t
r =
V/ C - 1 ^t = -r^t
(V/ C - 1 ). -1^t = r^t
RATIONALIZE THE ^t
^t√- V/ C + 1 = ^t√ r^t => (- V/ C)^1/t + 1 = r
L = Lo/x
x = 1/ √ 1 - v²/c²
Extends up to -v²/c²
v=
L = Lo/ 1/ √ 1 - v²/c²
INVERT 1/ √ 1 - v²/c²
L = Lo. √ 1 - v²/c²
L/Lo = √ 1 - v²/c²
ELIMINATE THE RATIO , by ²
T = √(ugx)²+ (To)²
√ extends to To)²
x=
√T² - To²/ug = x
√extends to To²
If f(x)=x²-6x and g(f(x))=-2x²+12x+5 what is the value of g(-9)?
1) f(x) = - 9 : x² - 6x = -9
x= 3
2) -2.3²+ 12.3 + 5 = -5 + 41 = 23
If f(x)= 7x -1 and f(g(x)) = 14x + 6, what equartion is equal g(x)
7 ( g(x) ) - 1 = f(g(x)) = 14x + 6
7. g(x) - 1 = 14x + 6
7. g(x) = 14x + 7
g(x) = 14x + 7 : 7 = 2x + 1
If f(x) = 5x +2 and f(g(x) ) = 5x² + 5x - 3 , what equation is equals to g(x)
f(g(x)) = 5.g(x) + 2 = 5x² + 5x - 3 5 g(x) = 5x² + 5x - 5 : 5 g(x) = x² + x - 1
If f(g(x) ) = 9x² and g(x)= 3x² - 2 . What equation is equals to f(x)
√3x+ √2
√3x + 2
3x+2
3x+6
SINCE ALL THE ANSWERS ARE LINEAR, DO IT LINEAR: f(x) = ax+ b
a ( 3x² -2 ) + b = 3x²a - 2a + b = 9x²
2) COMPARE
3x²a = 9x² -> a = 3
-2a+b = 0
3) a = 3, so b =
2.3 + b =0
6=b
With f (x-a) :
If f(x)=2x² +3 and f(x-a)=2x²-8x+11 what is the value of a?
1) Replace the “x” in f(x) by (x-a) :
2 (x-a)² + 3 = 2 ( x² -2ax + a² ) + 3 =>
2x² - 4ax + 2a² + 3
2) COMPARE with the f(x-a)
2x² - 4ax + 2a² + 3
2x²-8x+11
-4ax = -8x -> a = 2
2a² + 3 = 11
With g(x+a):
If ( g(x) ) = 1 / (x-2) (x+2) and g(x+a)= 1/ (x+2 ) (x+6) ;
a =
1) Replace x with x+a
1/ (x+a - 2 ) ( x+a + 2)
2) COMPARE
1/ (x+a - 2 ) ( x+a + 2) = 1/ (x+2 ) (x+6)
(x+a - 2 ) ( x+a + 2) = (x+2 ) (x+6)
a - 2 = 2 = 4
a + 2 = 6 = 4
a=4
h(x) = x+2/ x - 5 ; what equation is equivalent to h (h(x))
h(x) inside h(x) = replace h(x) in x value of h(x)
(x+2/ x - 5) + 2 / (x+2 / x - 5 ) - 5
2) MMC : x+2/ x - 5 + 2 and x+2/ x - 5 - 5
x+2 + 2 (x-5) = x+2 + 2x - 10 = 3x - 8
x+2+ -5 ( x-5) = x+2 - 5x + 25 = -4x +27
3) Go back to ratio form
3x-8/ -4x+27
Scatterplot graphs:
y = ax + b
Explain how to discover each letter
1) Draw a line from the most-left to the most right
a = slope : ( 0,y1 ) (x2,y2) = y2 - y1/ x2- 0
Isn’t obligatory to x=0. Wherever the line exactly pass through the points is valid
b = The first y-value that the line begins
or replace the values found in the a ( x1,y1) or (x2,y2) and replace in the equation to discover the b
! When the line in the graph is increasing => slope is + and if is exponential, the base is > 1 . EX: (1.05)^x
And vice-versa.
If 2+2(8347-4783) = -2j, how much is - j ?
SIMPLIFY the 2s
1 + 8347-4783 = -j
1 + 3564 = 3565 = -j
(3-4t) - 12 /2 < 2t + 3/2
(3-4t) - 12 is the numerator of 2
only 3 is the numerator of 2
Do not distribute “-“ to (3-4t)
THEY ARE SUBCTRATING :
-9 - 4t/2 < 2t + 3/2
Take out the denominator, by multiplying both sides -9 - 4t < 2.(2t + 3/2 ) -9 - 4t < 4t + 3 8t < 12 t < 3/2
WHAT IS THE VALUE OF B IF THERE ARE NO SOLUTIONS
“ “ “ “ INFINITIVE SOLUTIONS
10+7x=4x+2+bx
In the equation shown above, b is a constant. For what value of bbb does the equation have no solutions?
3+11x=cx+2(4x−1)+5
NO SOLUTIONS : The normal values should NOT be equal , therefore B = Ax
INFINITIVE SOLUTIONS: The normal values SHOULD BE equal
10+7x=4x+2+bx : 10+ 7x-4x = 2 + bx => 10 + 3x = 2 + bx ;
Since 10 is different 2 , bx = 3x = b = 3
3+11x=cx+2(4x−1)+5 : 11x - 8x = -2+ 5 + cx ;
Since 3 = 3 : 3x = cx = 3
4(80+n)=(3k)n
In the equation above, k is a constant. For what value of kkk are there no solutions to the equation?
NO SOLUTIONS = k should be equal to A.n
4n = (3k)n
4/3 = k -> Therefore, k.3 = 4
