math algebra 1 Help

Question

the sum of two numbers is 4 less than twice the larger number. the larger number decreased by the three times the smaller number is -20 Find the numbers?

And

17:00 in a piggy bank. he only has pennies and nickels. she has a total of 1140 coins. how many of each coin does he have?

and

tickets for the spring music are on presale on Friday there were 3 adults tickets and 9 students. tickets sold for $75. on Saturday there were 8 adult tickets and 5 student tickets old for 67. how much does each type of ticket cost?

Answer 1

Problem 1:

let x be the smaller number

let y be the larger number

then from problem statement x + y = 2y - 4 ...(1)

y - 3x = -20 .... (2)

solve for x in (1), add -y to both sides of the equation, x + y - y = 2y - y -4, the x = y -4 ...(3)

Substitute (3) in (2) , y - 3( y-4) = -20, expand, y - 3y + 12 = -20 , and -2y + 12 = -20

add -12 to both sides of the equation, -2y + 12 -12 = -20 -12, then -2y = -32

divide both sides by -2, then (-2y/-2)=(-32/-2), and y = 16

Substitute y =16 in (3) , then x = 16-4 = 12

Result: the larger number is y =16, and the smaller number is x=12

Problem 2:

let x be the number of pennies and y be the number of nickles

You have $17 in the piggy bank that is 1700 cents

if the penny is worth 1 cent and each nickel is worth 5 cents then you can write an equation that represent the total amount in the piggy bank as such:

1 cent ( number of pennies) + 5 cent ( number of nickles) = 1700 cents, in math form it will be:

1( x) + 5 (y) = 1700, or simply x + 5y = 1700 ...(1)

The 2nd equation comes from the problem statement that the combined number of pennies and nickles is 1140 or :

number of pennies + number of nickles = 1140, in math form it will be:

x + y = 1140...(2)

combining (1) and (2) together you have two equations two unknowns x + 5y = 1700 ...(1)

x + y = 1140...(2)

Multiply (2) by -1 and you get x + 5y = 1700 ...(1)

-x - y = -1140 ...(3)

Add (1) and (3) and you get 4y = 560 , Divide both sides by 4, (4y/4) = (560/4), and y = 140

Substitute y = 140 in (2), then x + 140 = 1140, add -140 to both sides, x + 140 -140 = 1140 -140,

and x = 1000

Check by substituting into (1), x + 5y = 1700 ...(1), 1000 + 5(140) = 1000 + 700 = 1700, checks

Result: you have 1000 pennies and 140 nickles in the piggy bank that totals 1700 cents or $17 dollars

Problem 3

Let x be the cost of an adult ticket

Let y be the cost of a student ticket

From the problem statement 3 adults and 9 students cost is $75 on Friday

You can write an equation in terms of cost as follows:

number of adults ( cost of an adult ticket )+ number of students (cost of a student ticket )= $75

3 x x + 9 x y = 75

3x + 9 y = 75 ...(1)

Similarly From the problem statement 8 adults and 5 students cost is $67 on Saturday

number of adults ( cost of an adult ticket )+ number of students (cost of a student ticket )= $67

8 x x + 5 x y = 67

8x + 5 y = 67...(2)

Combining the equations: 3x + 9y = 75 ...(1) Solve two equations two unknowns by elimination method

8x + 5y = 67...(2)

Multiply (1) by -8, and (2) by 3 and you get -8( 3x + 9y ) = -8 (75) ...(3)

3( 8x + 5y) = 3 (67) ...(4)

Then -24x -72y = -600 ...(3)

and 24x + 15y = 201...(4)

Add (3) and (4), -57y=-399, divide both sides by -57, ( -57y/-57) = (-399/-57), and y = 7

Substitute y = 7 in (1), then 3x+9(7)=75, 3x+63=75, add -63 to both sides, 3x+63-63=75-63, 3x=12,

divide both sides by 3, then (3x/3)=(12/3), and x = 4

Check by substituting the values of x and y in 92), 8(4)+5(7)=32+35=67, checks

Result: The cost of an adult ticket is $4 and the cost of a student ticket is $7, which really doesn't make much sense because the cost for the student should actually be less than the adult, but this is how the problem is formulated. LOL