Please please please help me with math problems!!!
1) The highest scorer of the women's basketball championship was Jessica Bradley. She scored 94 more points than Tina Harner, her teammate. Together, Bradley and Harner scored 1524 points. How many points did each plaayer score during the championship?
2)Jen Butler has been pricing Speed-Pass train fares for a group trip to New York. Three adults and four children must pay $125. Two adults and three children must pay $89. Find the price of the adult's ticket and price of a child's ticket.
1) The highest scorer of the women's basketball championship was Jessica Bradley. She scored 94 more points than Tina Harner, her teammate. Together, Bradley and Harner scored 1524 points. How many points did each plaayer score during the championship?
2)Jen Butler has been pricing Speed-Pass train fares for a group trip to New York. Three adults and four children must pay $125. Two adults and three children must pay $89. Find the price of the adult's ticket and price of a child's ticket.
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1) Jessica's score is x; Tina's score is y.
x=y+94 <-- Jessica's score may be written as Tina's score plus 94
x+y=1524 <-- Sum of their scores
__________
This is then solved as a system of linear equations.
y+94+y=1524 <-- instead of x in the second equation, a part of the first equation is inserted
2y=1524-94
2y=1430 /:2
y=715
x=715+94 <-- a value of y is inserted into the first equation
x=809
2) Price of a ticket for an adult is x, for children y.
3x+4y=125 <-- 3 adults + 4 children = 125
2x+3y=89 <-- 2 adults + 3 children = 89
__________
Another system of linear equations.
3x=125-4y <-- modifying the first equation to get only x on the left side
x=(125+4y)/3
2*(125-4y)/3 +3y=89 <-- x from the first equation, got in the row above, is inserted into the second one
250/3 - 8y/3 + 3y=89
-8y/3 + 3y = 89 - 250/3 <-- getting only y on the left side
(-8y+9y)/3 = (267-250)/3 <-- Addition of fractions
When multiplyed by 3 and calculated:
y=17
Now insert the value of y in an equation where only x is on the left side
x=(125-4*17)/,
x=57/3
x=19
x=y+94 <-- Jessica's score may be written as Tina's score plus 94
x+y=1524 <-- Sum of their scores
__________
This is then solved as a system of linear equations.
y+94+y=1524 <-- instead of x in the second equation, a part of the first equation is inserted
2y=1524-94
2y=1430 /:2
y=715
x=715+94 <-- a value of y is inserted into the first equation
x=809
2) Price of a ticket for an adult is x, for children y.
3x+4y=125 <-- 3 adults + 4 children = 125
2x+3y=89 <-- 2 adults + 3 children = 89
__________
Another system of linear equations.
3x=125-4y <-- modifying the first equation to get only x on the left side
x=(125+4y)/3
2*(125-4y)/3 +3y=89 <-- x from the first equation, got in the row above, is inserted into the second one
250/3 - 8y/3 + 3y=89
-8y/3 + 3y = 89 - 250/3 <-- getting only y on the left side
(-8y+9y)/3 = (267-250)/3 <-- Addition of fractions
When multiplyed by 3 and calculated:
y=17
Now insert the value of y in an equation where only x is on the left side
x=(125-4*17)/,
x=57/3
x=19
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1)
Using the 1st letter of the surnames:
Bradley's score B = H + 94
B + H = 1524
using second equation, substitute for either B or H in the first and solve.
Once you've found either B or H, work out the other using either equation.
2)
using A for adults and C for children
3A + 4C = 125
2A + 3C = 89
Again, you can solve these two simultaneous equations in similar way.
Try it!
Using the 1st letter of the surnames:
Bradley's score B = H + 94
B + H = 1524
using second equation, substitute for either B or H in the first and solve.
Once you've found either B or H, work out the other using either equation.
2)
using A for adults and C for children
3A + 4C = 125
2A + 3C = 89
Again, you can solve these two simultaneous equations in similar way.
Try it!
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Q1. let jessica & tina's goals be j&t respectively; now
j=94+t
i.e j-t=94 (a); also
j+t=1524 (b)
(a)+(b)
2j=1618
j=1618/2=809 goals
(a)-(b)
-2t=-1430
t=1430/2=715 goals
Q2. let adult & children be a & c respectively,thus
3a+4c=125 (a)
2a+3c=89 (b)
from (a) a=(125-4c)/3 (c)
putting this in (b) we get
2(125-4c)/3+3c=89
c=17 a=19
j=94+t
i.e j-t=94 (a); also
j+t=1524 (b)
(a)+(b)
2j=1618
j=1618/2=809 goals
(a)-(b)
-2t=-1430
t=1430/2=715 goals
Q2. let adult & children be a & c respectively,thus
3a+4c=125 (a)
2a+3c=89 (b)
from (a) a=(125-4c)/3 (c)
putting this in (b) we get
2(125-4c)/3+3c=89
c=17 a=19