Why is 2 equations with the same 2 variables required to solve for the variables?
- 1000 tickets were sold. Adult tickets cost $8.50, children's cost $4.50, and a total of $7300 was collected. How many tickets of each kind were sold?
- Mrs. B. invested $30,000; part at 5%, and part at 8%. The total interest on the investment was $2,100. How much did she invest at each rate?
- Samantha has 30 coins, quarters and dimes, which total $5.70. How many of each does she have?
- "36 gallons of a 25% alcohol solution"
means: 25%, or one quarter, of the solution is pure alcohol.
One quarter of 36 is 9. That solution contains 9 gallons of pure alcohol.
Here is the problem:
How many gallons of 30% alcohol solution and how many of 60% alcohol solution must be mixed to produce 18 gallons of 50% solution?
"18 gallons of 50% solution" means: 50%, or half, is pure alcohol. The final solution, then, will have 9 gallons of pure alcohol. - A saline solution is 20% salt. How much water must you add to how much saline solution, in order to dilute it to 8 gallons of 15% solution?
- It takes 3 hours for a boat to travel 27 miles upstream. The same boat can travel 30 miles downstream in 2 hours. Find the speeds of the boat and the current.
- A total of 925 tickets were sold for $5,925. If adult tickets cost $7.50, and children's tickets cost $3.00, how many tickets of each kind were sold?
- Mr. B. has $20,000 to invest. He invests part at 6%, the rest at 7%, and he earns $1,280 interest. How much did he invest at each rate?
- How many gallons of 20% alcohol solution and how many of 50% alcohol solution must be mixed to produce 9 gallons of 30% alcohol solution?
- 15 gallons of 16% disenfectant solution is to be made from 20% and 14% solutions. How much of those solutions should be used?
- It takes a boat 2 hours to travel 24 miles downstream and 3 hours to travel 18 miles upstream. What is the speed of the boat in still water, and how fast is the current?
- An airplane covers a distance of 1500 miles in 3 hours
when it flies with the wind, and in 3 1/3 hours when it flies against the wind. What is the speed of the plane in still air?
ok it is necessary to have two different equations to solve for the same two variables because if the two variables are unknown it is impossible to solve for both of them in one equation. So if two different equations are known representing the relationship of these variables the values can b substituted from one equation into the other to remove one variable and end up with a single equation consisting of one variable. Simultaneous equations are used to find the intersection of lines,curves or both.
ReplyDeleteThis comment has been removed by the author.
ReplyDelete1) in question one we let the number of adult tickets sold =x
ReplyDeleteand the number of children tickets sold=y
x+y=1000..............(1)
8.50x+4.50y=7300..............(2)
x=1000-y............(3)
sub (3) in (2)
8.50(1000-y)+ 4.50y=7300
8500-8.50y+4.50y=7300
-4y=7300-8500
-4y=-1200
Multiply both sides by -1
4y=1200
y=1200/4
y=300
Sum y=300 in (3)
x=1000-300
x=700
so 700 adult tickets were sold
and 300 children tickets were sold
the two varibale si connected by the addition or subtraction sign.
ReplyDeleteTwo varibles are needed so one varible could be cancelled of and solve the equation.
cap guy has done exactly what i was going to do in number 1. except, i wouldn't left the minus sign and cancel it out. so i would have to agree with your steps. also i found no errors.
ReplyDeleteno. 7
ReplyDeletelet x = adult ticket , y= child ticket
7.5x + 3y = 5925 ---------(1)
x + y = 925 ----------(2)
from (2)
y = 925 - x -------(3)
sub (3) into (1)
7.5x + 3(925 - x) = 5925
7.5x + 2775 -3x = 5925
4.5x = 5925 - 2775
4.5x=3150
/ throughout by 4.5; x= 700
sub x=700 in (3)
y= 925 - 700
y = 225
therefore ticket sold = 700 adult and 225 child
Question 3)Substitution Method:
ReplyDeletelet x=total number of quarters
let y=total number of dimes
x+y=30_______________(1)
0.25x+0.10y=5.70_____(2)
(1): x=30-y__________(3)
subst (3)into(2): 0.25(30-y)+0.10y=5.70
7.50-0.25y+0.10y=5.70
7.50-0.15y=5.70
-0.15y=5.70-7.50
-0.15y=-1.8
y= -1.8/-0.15
y=12
To find x subst y value into eqn (3), because it is the simplest eqn
x=30-y
x=30-12
therefore x=18
To conclude we can now state that:
Total number of quarters=18
Total number of dimes=12
Question 4)Elimination Method:
ReplyDeletelet x=total number of 30% alcohol solution gallons
let y=total number of 60% alcohol solution gallons
x+y=18_______________(1)
30x+60y=900__________(2) 50%*18gallons=900
(1)*30: 30x+30y=540__(3)
(2)-(3): 30y=360
y=360/30
y=12
To find x subst y value into eqn (1)
x+y=18
x+12=18
x=18-12
x=6
To conclude we can now state that:
total number of 30% alcohol solution gallons=6
total number of 60% alcohol solution gallons=12
Question 5)No specific Method required:
ReplyDeletelet x=total number of 0% saline solution gallons or total number of gallons of water
let y=total number of 20% saline solution gallons
x+y=8_______________(1)
0x+20y=120__________(2) 15%*8gallons=120
(2): 0x+20y=120
20y=120
y=120/20
y=6
To find x subst y value into eqn (1)
x+y=8
x+6=8
x=8-6
x=2
To conclude we can now state that:
x=total number of 0% saline solution gallons or total number of gallons of water=2
y=total number of 20% saline solution gallons=6
i agree with cap guy for the first one his explaination was fantas tic
ReplyDeleteno. 12
ReplyDeletespeed with wind =1500/3
=500
speed against wind
-1500/3 1/3
=450
therefore speed in still air=
500+450
-------
2
475 miles per hour
Firstly two variables are connected to signify that there is a comparison between the two and how these values pertain to the resultant of the equation. By putting two variables in one equation this makes it exceedingly difficult or impossible to solve but by using another equation with the same to variables you can now make one of the variables the subject of the formula and then substitute this equation into the other so that it would be in terms of only one variable. This method is known as the substitution method and it works under the postulation that you can for one variable (in terms of another) and plug in this value into the other equation an solve.
ReplyDeletethe 2 variables are connected since they are needed if there is only one then there would be no need for any calculations..the answer would be there.for example 4x=8 ,, x=2. if there are two variables in one equation then there would be need for calculations..
ReplyDeletetwo equations are needed since you cannot get the figures for the variables from one .. u use substitution or elimination method..
question(1)
ReplyDeleteadult tickets =$8.50
children tickets =$4.50
1000 sold
outcome=$7300
let a= adult tickets
c= children tickets
a + c = 1000................1
8.50a + 4.50c = 7300........2
c += 1000- a..........3
sub 3 into 2
8.50a + 4.50(1000 - a)= 73000
8.50a + 4500 -4.50 a = 7300
8.50a - 4.50a = 7300 - 4500
4a = 2800
a= 700 tickets
sub a = 700 into 3
c= 1000-700
c= 300 tickets
2 variables are connected because in life the are always more than one elements affecting the outcome of a situation, and you must take them both into consideration before you can calculate or determine the overall outcome
ReplyDeletetwo equations with the same two variables are needed so you can combine the two equations by elimination or substitution to get one equation that has only one of the variable(x), within that equation there'll be an expression that represents the other variable (y) just in another form. you're able to calculate what x is without having to deal with y but having an expression that has the same effect on the equation as y.
ReplyDelete2 variables are usually connected to show how they are related to each other.
ReplyDelete2 equations are required to solve for 2 variables because only using one equation can only allow for a relation between the two variables, for example, X+Y=5 is an equation, but none of the variables can be found from this equation, all we will be able to know from this if we transpose is the relationship between X and Y, where:
ReplyDeleteX=5-Y or Y=5-X.
However, In a simultaneous equation, there are 2 different representations of the same variables, for example:
X+Y=5 -(1)
X-Y= 12 - (2)
From those equations, both the values of X and Y can be found, as one equation can be used to find the relationship between X and Y, and the other can be used to find the values of X and Y
from my last comment:
ReplyDeleteX+Y = 5 -(1)
X-Y = 12 -(2)
Using the method of elimintation:
(1) + (2): X+Y + (X-Y) = 5+12
X+Y + X-Y = 17
X = 17
Substituting: X = 17 in (1)
17 + Y = 5
Y = 5-17
Y = -12
As can be seen,we were able to obtain the value of X and the value of Y
two equations are used to solve for two equations as you can make on of the variables the subject of one of the equations and then substitute this into the other equation. when u do this, you can now eliminate one variable by finding its value. then, you can substitute the value of the solved variable into the other equation and find the value of the other variable.
ReplyDeleteif there were three variables to be solved for, there would have to be three equations to solve it. if there were four variables, four equations would be needed etc...
ReplyDeletelet the number of adults' tickets = x
ReplyDeletelet the number of children = y
8.50x + 4.50y = 7300.......(1)
x + y = 1000 ..............(2)
from (2);
y = 1000 - x ..............(3)
sub 'y' int (1);
8.50x + 4.50(1000 - x) = 7300
8.50x + 4500 - 4.50x = 7300
4x + 4500 = 7300
4x = 7300 - 4500
4x = 2800
x = 2800 / 4
x = 700
sub 'x' into (1);
8.50(700) + 4.50y = 7300
5950 + 4.50y = 7300
4.50y = 7300 - 5950
4.50y = 1350
y = 1350 / 4.50
y = 300
therefore: the number of adult tickets sold = 700
the number of children's tickets sold =300
2 variables are connected by the means of which, you need to have a known to find for an unknown
ReplyDeletetwo equations with the same variable are required to solve for the variables by: using one to find for the other and then checking to make sure your answer is correct (by substituting your ans. in one of the equations)
ReplyDeletein simulataneous equations 2 variables are connected because one is dependent on the other which suggest if one variable is known the other can be found
ReplyDelete2 equations with the same 2 variables are used to solve for these variables because the both equations provide an equivalence relation for the variables. with this in mind i.e the both equations are equivalent, the variables can be substituted into its 'equivalent' to obtain the value for these variables. since the equations are simultaneously related the values of the variables are dependent on each other
ReplyDeletenumber (1)
ReplyDelete8.50=x
4.50=y
8.50+y=7300
x+4.50=7300
multpult by 4.50
38.25x=25550
x=25550/38.25=667.9
667.9+y=7300
y=7300-667.9=6632.1
i dont know if this is correct
ques(2)
ReplyDeleteinvest=$30,000 and interst on investment =$2,100
>therefore actual
invest after interest=$30,000-$2,100=$27,900
>total rate = 5%+8%=13%
invest at 5%= 5/13*27,900
= $10,730.80
invest at 8%=8/13*27,900
=17,169.23
ah yes the answer to question 11
ReplyDeletedownstream=24/2=12mph
upstream=18/3=6mph
speed of boat engine=(12+6)/2=9mph
speed of current=3mph
do you understand the scene
not sure if it is correct but feel free to correct me question 10
ReplyDeletex=20%
y=14%
x+y=15 1
20x+14y=250 2
eq'n 1*14
14x+14y=210 3
eq'n 2-3
6x=40
x=6.66
sub x in eq'n 1
6.66+y=15
y=15-6.66=8.34
that's your answer
Two varibles are connected simply because it would make no sense to solve a one varible equation simultaneously, when the equation can be solved using simple algebra.
ReplyDeleteFor instance 3y=21 using algebra
y= 21/3
y= 7
2x+7y=21 which is the connection of two variables.The values for x and y can be infinite eg. x=3.5,y=2 or x=2,y=1 or x=0,y=3 etc.Ultimately we cannot find a solution for this equation given what we have so more information is needed in order to solve this problem.
For the second question, the concept is the same as the first question.
ReplyDeleteFor 2 equations there can be an infinite number of solutions. The reason why 2 equations with the same varibles are used is because a common relationship must exsist in order to solve them.
It is not possible to solve 2 equations with differnt variables because there is no relationship existing between the equations.
For eg. 2x+y=3, 5x+7y=23
The relationship betweeb x an y is common in both equations.
However 2m+n=5, 31x+4y=45 There is no way these two equations can be equated, there is no connection...and therefore they cannot be solved
Solving two equations simultaneously means to find the common solution of both the equations, i.e., a solution which satisfies both the equations. Such a common solution, if it exists, can be shown to be unique.
ReplyDeleteThe following are methods used to solve simultaneous equations:
(a) Method of elimination
(b) Method of substitution.
(c) Graphical method
(d) Matrix method
Question 1.
ReplyDeleteLet adult tickets= x
Let children tickets= y
x+y= 1000-----(1)
8.5x+ 4.5y= 7300-----(2)
Solving simultaneously using substitution method:
From (1) x=1000-y----(3)
Sub equation(3) into equation (2)
8.5(1000-y)+ 4.5y= 7300
8500-8.5y+4.5y= 7300
-8.5y+4.5y= 7300-8500
-4y= -1200
y= -1200/-4
y= 300
sub y=300 into equation (3)
x= 1000-(300)
x= 700
Therefore the number of adult tickets sold were 700 and the number of children's tickets sold were 300
Question 2.
ReplyDeleteNow i am unsure if the solution for this problem is correct, so please identify errors if u wish.
NB:For the second equation because we are dealing with percentages we must put it over 100.
Let x be the amount of money she invested at 5%
Let y be the amount of money she invested at 8%
x+y= 30 000------(1)
5/100x + 8/100y= 2 100
0.05x + 0.08y= 2 100------(2)
From (1) x= 30 000-y------(3)
sub equation (3) into equation (2)
0.05(30 000-y)+0.08y= 2 100
1500-0.05y+0.08y= 2 100
-0.05y+0.08y= 2100-1500
0.03y= 600
y= 600/0.03
y= 20 000
sub y= 20 000 into equation (3)
x= 30 000- (20 000)
x= 10 000
Therefore Mrs B invested $10 000 at 5% rate and
$20 000 at 8% rate.
Question 3.
ReplyDeleteAssumeing Quarters= 25cents and
Dimes= 10 cents
25 cents of $1.00 is 0.25
10 cents of $1.00 is 0.1
Let x represent quarters
Let y reprsent dimes
x+y= 30-----(1)
0.25x+0.1y= 5.7 (2)
From equation (1) x=30-y-----(3)
Sub equation (3) in equation (2)
0.25(30-y)+0.1y=5.7
7.5-0.25y+0.1y= 5.7
-0.25y+0.1y=5.7-7.5
-0.15y= -1.8
y= -1.8/-0.15
y= 12
sub y=12 into equation (3)
x= 30-(12)
x= 18
Therefore Samatha has 18 Quarters and 12 Dimes
Miss, Question 4 looks very similar to question 5 in the "Exponent teaser".
ReplyDeleteAlthough in this question 4, you have given us more information i regret to say that i am still a bit lost...Please enlighten me about this question please...the percentage values are confusing.
Question 1
ReplyDeletex + y = 1000...1
8.50 + 4.50 = 7300...2
1 x 8.50
x + y = 1000 (x8.50)
8.50x + 8.50y = 8500...3
2 - 3
8.50x-8.50x + 4.50y - 8.50y = 7300 - 8500
4y = -1200
4y/4 = -1200/4
-y = -300 (x-1)
y = 300
sub y into 1
x + 300 = 1000
x + 300 - 300 = 1000 - 300
x = 700
soln(700,300)
number of $8.50 tickets sold = 700
number of $4.50 tickets sold = 300
x + y = 30...(1)
ReplyDelete1/4 + 1/10 = 5.70...(2)
(2) x 10
.25x + .1y = 5.70 (x10)
2.5x + 1y = 57...(3)
(2) - (3)
x - 2.5x + y - y = 30 - 57
-1.5x = -27
-1.5x/1.5 = -27/1.5
-x = -18 (x-1)
x = 18
sub x into (1)
18 + y = 30
18 - 18 + y = 30 - 18
y = 12
soln(18,12)
1) in question one we let the number of adult tickets sold =x
ReplyDeleteand the number of children tickets sold=y
x+y=1000..............(1)
8.50x+4.50y=7300..............(2)
x=1000-y............(3)
sub (3) in (2)
8.50(1000-y)+ 4.50y=7300
8500-8.50y+4.50y=7300
-4y=7300-8500
-4y=-1200
Multiply both sides by -1
4y=1200
y=1200/4
y=300
Sum y=300 in (3)
x=1000-300
x=700
so 700 adult tickets were sold
and 300 children tickets were sold
Question 7.
ReplyDeleteLet adult tickets= m
Let children tickets= n
m+n= 925-----(1)
7.5m+ 3n=5,925-----(2)
Solving simultaneously using substitution method:
From (1) n=925-m----(3)
Sub equation(3) into equation (2)
7.5m+ 3(925-m)= 5,925
7.5m+2,775-3m= 5,925
7.5m-3m= 5,925-2,775
4.5m= 3,150
m= 3,150/4.5
m=700
sub m=700 into equation (3)
n= 925-(700)
n= 225
Therefore the number of adult tickets sold were 700 and the number of children's tickets sold were 225
Question 8.
ReplyDeleteIf anyone wishes to correct me, please feel free.
Let e be the amount of money he invested at 6%
Let f be the amount of money he invested at 7%
e+f= 20,000-----(1)
0.06e+0.07f= 1,280-----(2)
From equation (1)
e= 20,000-f----(3)
Solving simultaneously through substitution
Sub equation (3) into equation (2)
0.06(20,000-f)+0.07f= 1,280
1,200-0.06f+0.07f= 1,280
-0.06f+0.07f= 1,280-1,200
0.01f= 80
f= 80/0.01
f= 8,000
sub f= 8,000 into equation (3)
e = 20,000-8,000
e = 12,000
Therefore Mr B invested $12 000 at 6% rate and $8 000 at 7% rate.
for ques 1
ReplyDeletelet the number of adults tickets = x
let the number of children tickets = y
8.50x + 4.50y = 7300 eq1
x + y = 1000 eq2
from eq 2
y = 1000 - x eq3
sub y in eq1
8.50x + 4.50(1000 - x) = 7300
8.50x + 4500 - 4.50x = 7300
4x + 4500 = 7300
4x = 7300 - 4500
4x = 2800
x = 2800 / 4
x = 700
sub x in eq1
8.50(700) + 4.50y = 7300
5950 + 4.50y = 7300
4.50y = 7300 - 5950
4.50y = 1350
y = 1350 / 4.50
y = 300
the number of adult tickets sold = 700
the number of childrens tickets sold =300
1)adult tickets= x
ReplyDeletechildren tickets=y
x+y=1000..............(1) 8.50x+4.50y=7300..............(2)
(1)
x=1000-y............(3)
sub (3) in (2)
8.50(1000-y)+ 4.50y=7300
8500-8.50y+4.50y=7300
-4y=7300-8500
-4y=-1200
y=1200/4
=300
Sub y=300 in (3)
x=1000-300
x=700
invest=$30,000
ReplyDeleteinterst =$2,100
invest befor interest=$30,000-$2,100=$27,900
total % = 5%+8%
=13%
therefore 5/13=0.385
invest for 5% = 0.385 * 27,900
= $10,730.80
therefore invest for 8% = $27900 - $10730.8
=$17169.2
7) x = adult ticket
ReplyDeletey= children ticket
x + y = 925 ----------(1)
7.5x + 3y = 5925 ---------(2)
(1)
y = 925 - x -------(3)
sub (3) into (2)
7.5x + 3(925 - x) = 5925
7.5x + 2775 -3x = 5925
4.5x = 5925 - 2775
4.5x=3150
x=3150/4.5
x= 700
y= 925 - 700
y = 225
two equations is necessary to solve for two variables . since two equations are used representing the relationship of the variables the values can be substituted from one equation into the other to remove one variable and end up with a single equation consisting of one variable.
ReplyDeletewhich can then be found by transposing makin the unknown the subject of the formular.
ques one.
ReplyDeleteadults = A
children = c
total tickets sold is equal to 1000
so a+c=1000--------(1)
adults = $8.50
children = $4.50
$8.50a + $4.50c = $7300-------(2)
so from eq'n (1)
a=1000-c-----------------(3)
so therefore
$8.50(1000-c) + $4.50c = $7300
8500 - $8.50c + $4.50c = $7300
8500 - $7300 = $8.50c - $4.50c
1200 = $4c
300 = c
subst c = 300 into eq'n (1)
a + c =1000
a + 300 =1000
a = 1000 - 300
a = 700
adults tickets sold = 700
children tickets sold = 300
it is necessary to have two different equations to solve for the same two variables because if the two variables are unknown it is impossible to solve for both of them in one equation because we cannot find two unknowns
ReplyDeleteusing simultaneous equations to solve the following:
ReplyDeletechildren tickets =$4.50
adult tickets =$8.50
total tickets sold = 1000
total $ earn =$7300
let a = adult tickets
c = children tickets
a + c = 1000--------------1
8.50a + 4.50c = 7300------2
c + = 1000- a--------------3
subtitiute eq'n 3 into eq'n 2
8.50a + 4.50(1000 - a)= 73000
8.50a + 4500 -4.50 a = 7300
8.50a - 4.50a = 7300 - 4500
4a = 2800
a= 700 tickets
amount of adults tickets sold =700
sub a = 700 into eq'n 3
c= 1000-700
c= 300 tickets
amount of children tickets sold =300
if two different equations are used to represent the relationship of 2 variables the values can b substituted from one equation into the other to eliminate one variable and end up with a single equation consisting of one variable. then one of the two variables can be solved and that value can be substuted into the original eq'n and find for the other variabl
ReplyDelete(2)
ReplyDeletex+ y= 30 (1)
.25x + .10y= 5.70 (2)
from (2)(x= 30- y) (3)
sub (3) into (2)
.25(30- y) + .10y= 5.70
7.5- .25y + .10y= 5.70
7.5- .15y= 5.70
7.5- 5.70= .15y
1.8= .15y
1.8/.15=y
y=12
sub y=12 into (3)
x=30 -12
x=18
therefore she has 12 ten cents and 18 twenty five cents..
12) speed in the wind =1500/3
ReplyDelete=500
speed against wind =1500/3 1/3
=450
the total spped in the air is
500+450= 950
the average speed=950/2
=475 miles per sec
question 1
ReplyDeletex+y=1000..............(1)
8.50x+4.50y=7300..............(2)
x=1000-y............(3)
sub (3) in (2)
8.50(1000-y)+ 4.50y=7300
8500-8.50y+4.50y=7300
-4y=7300-8500
-4y=-1200
Multiply both sides by -1
4y=1200
y=1200/4
y=300
Sum y=300 in (3)
x=1000-300
x=700
so 700 adult tickets were sold
and 300 children tickets were sold
question 2
ReplyDeleteinvest=$30,000 and interst on investment =$2,100
>therefore actual
invest after interest=$30,000-$2,100=$27,900
>total rate = 5%+8%=13%
invest at 5%= 5/13*27,900
= $10,730.80
invest at 8%=8/13*27,900
=17,169.23
Question 3
ReplyDeleteSubstitution Method:
let x=total number of quarters
let y=total number of dimes
x+y=30_______________(1)
0.25x+0.10y=5.70_____(2)
(1): x=30-y__________(3)
subst (3)into(2): 0.25(30-y)+0.10y=5.70
7.50-0.25y+0.10y=5.70
7.50-0.15y=5.70
-0.15y=5.70-7.50
-0.15y=-1.8
y= -1.8/-0.15
y=12
To find x subst y value into eqn (3), because it is the simplest eqn
x=30-y
x=30-12
therefore x=18
To conclude we can now state that:
Total number of quarters=18
Total number of dimes=12
Question 4
ReplyDeleteElimination Method:
let x=total number of 30% alcohol solution gallons
let y=total number of 60% alcohol solution gallons
x+y=18_______________(1)
30x+60y=900__________(2) 50%*18gallons=900
(1)*30: 30x+30y=540__(3)
(2)-(3): 30y=360
y=360/30
y=12
To find x subst y value into eqn (1)
x+y=18
x+12=18
x=18-12
x=6
To conclude we can now state that:
total number of 30% alcohol solution gallons=6
total number of 60% alcohol solution gallons=12
Try solving simultaneous quadratic equations below.
ReplyDeletea)
x^2-4x=2=y
x = y + 4
x+y=1000 tickets(x bieng the # of adults and y is the # of children)
ReplyDelete8.5x+4.5y=7300
x+y=1000 eq1
8.5x+4.5y=7300 eq2
x=1000-y eq3
sub eq3 into eq2
8.5(1000-y)+4.5y=7300
8500-8.5y+4.5y=7300
8500-4y=7300
8500-7300=4y
1200=4y
1200/4
300=y
sub 300=y into eq1
x+300=1000
x=1000-300
x=700
for quetion 12
ReplyDeletewind speed
=1500/3
=500
speed against wind
-1500/3 1/3
=450
therefore speed in still air
=500+450
-------
2
475 miles per hour
for question 11
ReplyDelete24/2=12mph downsream
18/3=6mph upstream
boat engine speed=(12+6)=18/2=9mph
speed of current=3mph
question 7
ReplyDeletex = adult, y= child
7.5x + 3y = 5925 eq1
x + y = 925 eq2
from eq2
y = 925 - x eq3
sub eq3 into eq1
7.5x + 3(925 - x) = 5925
7.5x + 2775 -3x = 5925
4.5x = 5925 - 2775
4.5x=3150
x=3150/4.5
x=700
sub x=700 in eq3
y= 925 - 700
y = 225
adult=700, children=225
qeustion 3
ReplyDeletex=number of quarters
y=number of dimes
x+y=30 eq1
0.25x+0.10y=5.70 eq2
from eq1
x=30-y eq3
subst eq3 into eq2
0.25(30-y)+0.10y=5.70
7.50-0.25y+0.10y=5.70
7.50-0.15y=5.70
-0.15y=5.70-7.50
-0.15y=-1.8
y=-1.8/-0.15
y=12
sub y value into eq3
x=30-y
x=30-12
x=18
quarters=18
dimes=12
Two varibales are connected by an addition or a subtraction sign.
ReplyDelete.
Two(2) varibles are needed in an equation in order for one varible to be cancelled of to solve the equation
ReplyDeletex^2-4x=2=y
ReplyDeletex = y + 4
something wrong here
solve the simultaneous equation
ReplyDelete3x + y= 7
5x - 3y= 7
solve the simultaneous equation by substition method
ReplyDelete2x + 3y = 2
x - 5y = 14
the bill for 3 cups of mango bubble tea and 4 cups of green tea bubble tea is $13.40 while the bill for 5 cups mango bubble tea and 2 cups of green tea bubble tea is $13.00 find the cost of each cup of mango bubble tea and each cup of green tea bubble tea.
ReplyDeletesolve the following simultaneous equation by elimination method
ReplyDeletex + 5y = 2
2x - 5y = 19
One packet of biscuits costs $x and one cup of ice-cream cost $y.
ReplyDeleteone packet of buscuits and two cups of ice-cream cost $8.00
(i) write a pair of simultaneous equation in x and y to represent the given information above
(ii) solve the equation obtained in (i) above to find the cost of one packet of buscuits and the cost of one cup pf ice-cream
solve the following simultaneous equation by both substition an elimination method
ReplyDelete2x + 7y = 11
2x + 3y = 7
A stadium has two sections A and B
ReplyDeleteTickets for section A costs $a each
Tickets for section B costs $b each
Johanna paid $105. for 5 section A tickets and 3 section B tickets.
Raiyah paid $63. for 4 section A tickets and 1 section B ticket.
(i) Write 2 equations in a and b to represent the information above.
(ii) Calculate the values of a and b.
x+y= 1000-----(1)
ReplyDelete8.5x+ 4.5y= 7300-----(2)
Solving simultaneously using substitution method:
From (1) x=1000-y----(3)
Sub equation(3) into equation (2)
8.5(1000-y)+ 4.5y= 7300
8500-8.5y+4.5y= 7300
-8.5y+4.5y= 7300-8500
-4y= -1200
y= -1200/-4
y= 300
sub y=300 into equation (3)
x= 1000-(300)
x= 700
adult tickets= x
ReplyDeleteLet children tickets= y
for number 1.
make this equ. 1.
ReplyDeletex+ y= 30 (1)
.25x + .10y= 5.70 (2)
from (2)(x= 30- y) (3)
sub (3) into (2)
.25(30- y) + .10y= 5.70
7.5- .25y + .10y= 5.70
7.5- .15y= 5.70
7.5- 5.70= .15y
1.8= .15y
1.8/.15=y
y=12
sub y=12 into (3)
x=30 -12
x=18
25 cents of $1.00 is 0.25
ReplyDelete10 cents of $1.00 is 0.1
Let x represent quarters
Let y reprsent dimes
x+y= 30-----(1)
0.25x+0.1y= 5.7 (2)
From equation (1) x=30-y-----(3)
Sub equation (3) in equation (2)
0.25(30-y)+0.1y=5.7
7.5-0.25y+0.1y= 5.7
-0.25y+0.1y=5.7-7.5
-0.15y= -1.8
y= -1.8/-0.15
y= 12
sub y=12 into equation (3)
x= 30-(12)
x= 18
no. 12
ReplyDeletespeed with wind =1500/3
=500
speed against wind
-1500/3 1/3
=450
therefore speed in still air=
500+450
-------
2
475 miles per hour
question 7
ReplyDeletex=no of adult tickets
y= no of children tickets
7.50x + 3y = 5925---------(1)
x + y = 925 --------------(2)
eq'n (1)
y = 925 - x ------------(3)
sub (3) into (2)
7.50x + 3(925 - x)= 5925
7.50x + 2775 - 3x = 5925
7.50x - 3x = 5925 - 2775
4.50x = 3150
x = 3150/4.5
x = 700
no of adult tickets sold; 700
no of children tickets (y);
y = 925 - x
y = 925 - 700
y = 225
let adult tickets= x
ReplyDeleteLet children tickets= y
x+y= 1000-----(1)
8.5x+ 4.5y= 7300-----(2)
Solving simultaneously using substitution method:
From (1) x=1000-y----(3)
Sub equation(3) into equation (2)
8.5(1000-y)+ 4.5y= 7300
8500-8.5y+4.5y= 7300
-8.5y+4.5y= 7300-8500
-4y= -1200
y= -1200/-4
y= 300
sub y=300 into equation (3)
x= 1000-(300)
x= 700
Therefore the number of adult tickets sold were 700 and the number of children's tickets sold were 300
adult tickets= x
ReplyDeletechildren tickets= y
x+y= 1000 (1)
8.5x+ 4.5y= 7300 (2)
From (1) x=1000-y (3)
Sub equation(3) into equation (2)
8.5(1000-y)+ 4.5y= 7300
8500-8.5y+4.5y= 7300
-8.5y+4.5y= 7300-8500
-4y= -1200
y= -1200/-4
y= 300
sub y=300 into equation (3)
x= 1000-(300)
x= 700
therefore x = 700 and y =300
hmmm ques.2 i not really sure how to handle this one. duncy_lion well i looked at your solution and i think i understand a little more now.
ReplyDeletetotal number of quarters = x
ReplyDeletetotal number of dimes = y
x+y=30 (1)
0.25x+0.10y=5.70 (2)
From eq (1)x=30-y (3)
subst (3)into(2)
0.25(30-y)+0.10y=5.70
7.50-0.25y+0.10y=5.70
7.50-0.15y=5.70
-0.15y=5.70-7.50
-0.15y=-1.8
y= -1.8/-0.15
y=12
subst y value into eqn (3)
x=30-y
x=30-12
x=18
Therefore y =12 and x 18
some simultaneous eq to slove please.
ReplyDelete(1) 2y= x + 5 -------(1)
y= 2x +10 -------(2)
(2) y= 2x^2 -4x +2 -------(1)
ReplyDeletey^2 = 4-2x --------(2)
(3) y = x +1 ---(1)
ReplyDeletex^2= y + 2 -----(2)
(4) y = 5x + x^2 -----(1)
ReplyDeletey = x^2 + 3x +2 ----(2)
two equations with the same two variables are necessary to find the value of the two variables since it is highly impossible to solve for the value of these two variables if you only had 1 equation...thus the two equations linking the two variables are necessary to solve for the variables.
ReplyDelete1.1000 tickets were sold. Adult tickets cost $8.50, children's cost $4.50, and a total of $7300 was collected. How many tickets of each kind were sold?
ReplyDeletelet x = the number of adult tickets sold
let y = the number of children tickets sold
x + y = 1000.......................(1)
8.50x + 4.50y = 7300...............(2)
from equation (1) x = 1000 - y.....(1a)
sub. equation (1a) into equation (2)
8.50(1000 - y) + 4.50y = 7300
8500 - 8.50y + 4.50y = 7300
8500 - 4y = 7300
8500 - 7300 = 4y
1200 = 4y
y = 1200/4
y = 300
when y = 300
x = 1000 - 300
x = 700
so 300 children tickets were sold and 700 adult tickets were sold.
3.Samantha has 30 coins, quarters and dimes, which total $5.70. How many of each does she have?
ReplyDeletea quarter is 0.25 of a dollar
a dime is 0.10 of a dollar
let x = the number of quarters
let y = the number of dimes
x + y = 30.....................(1)
0.25x + 0.10y = $5.70..........(2)
from (1) x = 30 - y............(1a)
sub. (1a) into (2)
0.25(30 - y) + 0.10y = 5.70
7.5 - 0.25y + 0.10y = 5.70
7.5 - 0.15y = 5.70
7.5 - 5.70 = 0.15y
1.8 = 0.15y
y = 1.8/0.15
y = 12
when y = 12
x = 30 - 12
x = 18
so she has 12 dimes (10 cents) and 18 quarters (25 cents)
(5) 4y= x + 11 ----------(1)
ReplyDeletey^2 = 2x + 7 --------(2)