the positive integer a b c satisfy a^2 + b -c is 100 and a+ b^2 - c is 124 what are they

We are given two equations involving positive integers \( a \), \( b \), and \( c \):

1. \( a^2 + b - c = 100 \)
2. \( a + b^2 - c = 124 \)

Our goal is to find the values of \( a \), \( b \), and \( c \).

### Step 1: Express \( c \) from both equations
From equation (1):
\[
c = a^2 + b - 100
\]
From equation (2):
\[
c = a + b^2 - 124
\]

Now we can set these two expressions for \( c \) equal to each other:
\[
a^2 + b - 100 = a + b^2 - 124
\]

### Step 2: Simplify the equation
Rearrange terms to move everything to one side:
\[
a^2 - a + b - b^2 = -124 + 100
\]
\[
a^2 - a + b - b^2 = -24
\]
This is the equation we need to solve.

### Step 3: Try values for \( a \) and \( b \)
Let's try different values of \( a \) and \( b \), as the equation is quadratic and involves integers.

#### Try \( a = 5 \):
\[
5^2 - 5 + b - b^2 = -24
\]
\[
25 - 5 + b - b^2 = -24
\]
\[
20 + b - b^2 = -24
\]
\[
b - b^2 = -44
\]
\[
b^2 - b - 44 = 0
\]
Solve this quadratic equation using the quadratic formula:
\[
b = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-44)}}{2(1)}
\]
\[
b = \frac{1 \pm \sqrt{1 + 176}}{2}
\]
\[
b = \frac{1 \pm \sqrt{177}}{2}
\]

Since \( \sqrt{177} \) is not an integer, \( a = 5 \) does not work.

#### Try \( a = 6 \):
\[
6^2 - 6 + b - b^2 = -24
\]
\[
36 - 6 + b - b^2 = -24
\]
\[
30 + b - b^2 = -24
\]
\[
b - b^2 = -54
\]
\[
b^2 - b - 54 = 0
\]
Solve this quadratic equation using the quadratic formula:
\[
b = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-54)}}{2(1)}
\]
\[
b = \frac{1 \pm \sqrt{1 + 216}}{2}
\]
\[
b = \frac{1 \pm \sqrt{217}}{2}
\]

Again, \( \sqrt{217} \) is not an integer, so \( a = 6 \) does not work.

#### Try \( a = 7 \):
\[
7^2 - 7 + b - b^2 = -24
\]
\[
49 - 7 + b - b^2 = -24
\]
\[
42 + b - b^2 = -24
\]
\[
b - b^2 = -66
\]
\[
b^2 - b - 66 = 0
\]
Solve this quadratic equation using the quadratic formula:
\[
b = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-66)}}{2(1)}
\]
\[
b = \frac{1 \pm \sqrt{1 + 264}}{2}
\]
\[
b = \frac{1 \pm \sqrt{265}}{2}
\]

Since \( \sqrt{265} \) is not an integer either, \( a = 7 \) does not work.

#### Try \( a = 8 \):
\[
8^2 - 8 + b - b^2 = -24
\]
\[
64 - 8 + b - b^2 = -24
\]
\[
56 + b - b^2 = -24
\]
\[
b - b^2 = -80
\]
\[
b^2 - b - 80 = 0
\]
Solve this quadratic equation using the quadratic formula:
\[
b = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-80)}}{2(1)}
\]
\[
b = \frac{1 \pm \sqrt{1 + 320}}{2}
\]
\[
b = \frac{1 \pm \sqrt{321}}{2}
\]

Since \( \sqrt{321} \) is also not an integer, \( a = 8 \) does not work.

At this point, trying different values

The positive integer a, b, c satisfy a^2 + b - c = 100 and a + b^2 - c = 124. What are they?

Explore how to solve a system of equations involving quadratic expressions: a^2 + b - c = 100 and a + b^2 - c = 124. Follow the step-by-step solution using algebraic techniques and the quadratic formula. Suitable for students in Grades 9-11.

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