AA = $1
AAA= $0.75
AA + AAA = 42
$1AA + $0.75AAA= $37
AA + AAA = 42
AA + AAA-AAA= 42- AAA
AA = 42- AAA
$1(42- AAA) + $0.75AAA= $37
$42 - AAA +0.75AAA = $37
$42 -0.25AAA= $37
$42-$42 -0.25AAA= $37 -$42
-0.25AAA= -5
-0.25AAA/-0.25 = -5/-0.25
AAA= 20
AA + AAA= 42
AA + 20 = 42
AA +20 -20 = 42-20
AA= 22
Check
$1AA + $0.75AAA= $37
$1(22)+ $0.75(20)= $37
$22 + $15 =$37
$37 = $37
Answer:
x = 17 and y = 37
Step-by-step explanation:
Let the numbers be x and y.
y = 3 + 2x
xy = 629
substituting y in the second equation we get
x(3+2x) = 629
3x + 2
= 629
2 + 3x - 629 = 0
Solving a quadratic equation by using its formula:
x = [-3 ±√(9-4(2)(-629)] / 4
x = [-3 ± √5041] / 4
x = [-3 ± 71] / 4
x = 17 and 
Ignore the negative number in x and just use x = 17 in the first equation we get
y = 3 + 2(17) = 37
Answer:
Step-by-step explanation:
There isn't one. The first term is x^4.
The other two are stated in terms of k. There is nothing common here.
You can do 2 of the three. Terms 2 and 3 have a common factor of 4k, so what you can get is
x^4 - 4k(k^2 - 1)
Answer:
Cost of small box of oranges = 7
Cost of small box of oranges = 13
Step-by-step explanation:
Step 1) Let the cost of small box of oranges = x
Let the cost of small box of oranges = y
Step 2)
Equation 1) Matt sells 3 small boxes and 14 large boxes for Php. 203
3x + 14y = 203 -------------------(I)
Equation 2) Ming sells 11 small boxes and 11 large boxes for Php. 220
11x + 11y = 220 ----------------(II)
Step 3:
Multiply equation (I) by 11 and equation (II) by (-3).
(I)*11 33x + 154y = 2233
(II)*(-3) <u>-33x - 33y = -660 </u> {Now add and x will be eliminated}
121y = 1573
y = 1573/121
y = 13
Substitute y = 13 in equation (II)
11x + 11*13 = 220
11x + 143 = 220
11x = 220 - 143
11x = 77
x = 77/11
x = 7
