Problem 1 :
Solve by elimination method.
3x + 4y = -25
2x - 3y = 6
Problem 2 :
Solve by elimination method
2x + 3y = 5
3x + 4y = 7
Problem 3 :
A park charges $10 for adults and $5 for kids. How many many adults tickets and kids tickets were sold, if a total of 548 tickets were sold for a total of $3750 ?
Problem 1 :
Solve by elimination method.
3x + 4y = -25
2x - 3y = 6
Solution :
3x + 4y = -25 ---- (1)
2x - 3y = 6 ---- (2)
Both x terms and y terms have different coefficients in the above system of equations.
Let's try to make the coefficients of y terms equal.
To make the coefficients of y terms equal, we have to find the least common multiple 4 and 3.
The least common multiple of 4 and 3 is 12.
Multiply the first equation by 3 in order to make the coefficient of y as 12 and multiply the second equation by 4 in order to make the coefficient of y as -12.
(1) ⋅ 3 ----> 9x + 12y = -75
(2) ⋅ 4 ----> 8x - 12y = 24
Now, we can add the two equations and eliminate y as shown below.
Divide each side by 17.
x = -3
Substitute -3 for x in (1).
(1)----> 3(-3) + 4y = -25
-9 + 4y = -25
Add 9 to each side.
4y = -16
Divide each side by 4.
y = -4
So, the values of x and y are -3 and -4 respectively.
Problem 2 :
Solve by elimination method
2x + 3y = 5
3x + 4y = 7
Solution :
2x + 3y = 5 ----(1)
3x + 4y = 7 ----(2)
Both x terms and y terms have different coefficients in the above system of equations.
Let's try to make the coefficients of x terms equal.
To make the coefficients of y terms equal, we have to find the least common multiple 2 and 3.
The least common multiple of 2 and 3 is 6.
Multiply the first equation by 3 in order to make the coefficient of x as 6 and multiply the second equation by -2 in order to make the coefficient of x as -6.
(1) ⋅ 3 ----> 6x + 9y = 15
(2) ⋅ -2 ----> -6x - 8y = -14
Now, we can add the two equations and eliminate x as shown below.
Substitute 1 for y in (1).
(1)----> 2x + 3(1) = 5
2x + 3 = 5
Subtract 3 from each side.
2x = 2
Divide each side by 2.
x = 1
So, the values of x and y are 1 and 1 respectively.
Problem 3 :
A park charges $10 for adults and $5 for kids. How many many adults tickets and kids tickets were sold, if a total of 548 tickets were sold for a total of $3750 ?
Solution :
Step 1 :
Let "x" be the number of adults tickets and "y" be the number of kids tickets.
No. of adults tickets + No. of kids tickets = Total
x + y = 548 ----(1)
Step 2 :
Write an equation which represents the total cost.
Cost of "x" no. adults tickets = 10x
Cost of "y" no. of kids tickets = 5y
Total cost = $3750
Then, we have
10x + 5y = 3750
Divide both sides by 5.
2x + y = 750 ----(2)
Step 3 :
Solve (1) and (2) using elimination method.
x + y = 548 ----(1)
2x + y = 750 ----(2)
In the above two equations, y is having the same coefficient, that is 1.
Multiply the first equation by -1 to get the coefficient of -1. And keep the second equation as it is.
Then, we have
-x - y = - 548
2x + y = 750
We can add the above two equations and eliminate y.
Then, we have
x = 202
Step 4 :
Substitute 202 for x in the first equation.
(1)----> 202 + y = 548
Subtract 202 from each side.
y = 306
So, the number of adults tickets sold is 202 and the number of kids tickets sold is 346.
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