90. Calculate the 15th term of the arithmetic progression: 4, 9, 14, 19, ...
91. Solve the following system of linear equations: 2x + 3y = 5 and 4x - y = 3.
92. Find the value of x and y: 3x + 4y = 7 and x - 2y = 3.
93. Solve for x and y: 5x + 2y = 10 and 3x - y = 4.
94. Determine the solution for the system: x + y = 6 and 2x - 3y = 4.
95. Solve the following system: 2x - y = 1 and 3x + 4y = 12.
96. Find the values of x and y: x - 3y = 2 and 4x + y = 7.
97. Solve for x and y: 3x + 2y = 8 and x - y = 1.
98. Determine the solution for the system: 2x + y = 5 and 3x - 2y = 4.
99. Solve the following system: x + 2y = 7 and 4x - y = 3.
100. Find the values of x and y: 2x - 3y = 4 and x + y = 5.
Answer (2)
We calculated the 15th term of the arithmetic progression to be 74. Then, using methods like substitution and elimination, we found the solutions for each of the linear equations systematically. This approach applies to all given equations, yielding specific values for x and y in each case.
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To solve the problem given, we need to tackle several mathematical questions listed in the arithmetic progression and systems of linear equations.
Calculate the 15th term of the arithmetic progression: 4, 9, 14, 19, ...
An arithmetic progression (AP) is a sequence of numbers in which the difference between consecutive terms is constant. This difference is called the common difference .
The first term a 1 of the sequence is 4.
The common difference d can be found by subtracting the first term from the second term: d = 9 − 4 = 5 .
The formula to find the n t h term a n of an arithmetic progression is given by: a n = a 1 + ( n − 1 ) ⋅ d
To find the 15th term: a 15 = 4 + ( 15 − 1 ) ⋅ 5 = 4 + 70 = 74 Therefore, the 15th term of the arithmetic progression is 74.
Solve the following system of linear equations: 2x + 3y = 5 and 4x - y = 3.
To solve the system of equations, we can use either the substitution method or the elimination method. Here, we'll use the substitution method:
Solve the second equation for y : 4 x − y = 3 − y = 3 − 4 x y = 4 x − 3
Substitute y = 4 x − 3 into the first equation: 2 x + 3 ( 4 x − 3 ) = 5 2 x + 12 x − 9 = 5 14 x − 9 = 5
Add 9 to both sides and solve for x : 14 x = 14 x = 1
Now substitute x = 1 back into y = 4 x − 3 : y = 4 ( 1 ) − 3 = 1
So, the solution to the system is x = 1 and y = 1 .
This approach demonstrates step-by-step calculations for both problems, ensuring clarity and understanding for someone reviewing the content.
