8th grade math problems and answers is the page, which gives all support to students of 8th grade.  Here you can get help with math problems.  The main highlight of these answers is that they have step by step explanation.

Following are few of the topic we cover in the 8th grade math.  You may enter any of the questions in 8th grade curriculum and get benefited.

• Decimals
• Fractions
• Algebraic expressions
• Integers
Following are the simple word problems solved by our tutors
Example 1: -
The sum of the square of the squares of two consecutive positive number is 61.  Find the numbers?
Solution: -
Given that the numbers are consecutive.  So let us assume that the numbers are x and x + 1.
Given that the sum of the square of the squares of two consecutive positive number is 61 That is x2 + (x + 1)2 = 61
x2+ x2+ 2x + 1 = 61
2x2 + 2x + 1 = 61
Subtracting 61 from both sides, we get
2x2+ 2x - 60 = 0
Dividing both sides by 2, we get
x2+ x - 30 = 0
We split the middle term
x2 + 6x - 5x - 30 = 0
Now grouping the first two and last two terms we get
x(x + 6) -5 (x + 6) = 0
taking x + 6 as common factor, we get
(x + 6)(x - 5) = 0
So x = -6 or 5
Since our integers are positive, x cannot be –6
So x = 5
When x = 5, (x + 1) = 5 + 1 = 6
So the required integers are  5 and 6
Example 2: -
Jack buys eight books for $44. Paperback books cost$4 and hardback cost $8. How much of each type did he buy? Solution: - Let x be the number of paperback books and y be the number of hardback books he bought.Given that buys eight books for$44.
That is x + y = 8…(1)
Also given that Paperback books cost $4 and hardback cost$8.
That is 4x + 8y = 44 ...(2)
Solving the two equations we get the required answer.
For this first we divide the second equation by 4, we get
x + 2y = 11…(3)
Subtracting equation (1) from (3) we get
y = 3
Substituting in equation (1), we get
x  + 3 = 8
x = 8 – 3 = 5
Therefore number of paperback books he bought = 5
Number of hardback books he bought = 3.
Example 3: -
Find the equation of the circle passing through the points (1,1), (2, -1) and (3, 2).
Solution: -
Let the required equation of the circle be x2+ y2 + 2gx + 2fy + c = 0...(1)
Since (1) passes through the given points, substituting the coordinates of the three points successively in (1), we get
2g + 2f + c = -2...(2)
4g - 2f + c = -5...(3)
6g + 4f + c = -13...(4)
Subtracting (2) from (3), we get
2g - 4f = -3...(5)
Subtracting (3) from (4), we get
2g + 6f = -8...(6)
Subtracting (5) from (6), we get
10f = -5
f = -5/10 = -1/2
Substituting f = -1/2 in (5), we get
2g - 4(-1/2) = -3
2g + 2 = -3
2g = -5
g = -5/2
Substituting the values of g and f in (2) we get
2(-5/2) + 2(-1/2) + c = -2
-5 - 1 + c = -2
-6 + c = -2
c = 4
Therefore the required equation is x2 + y2 - 5x - y + 4 = 0