Answer:
we do not have the integer coeficients, so the solutions may not be in the range -10 to 10 (it does not happen for all coeficients) so let's do other thing:
Ok, suppose you have the equations:
1) A*x + B*y = C
2) a*x + b*y = c
where A, B, C, a, b and d are knowed.
Let's solve them in the most general way, and you always can use those solutions in the future, just need to replace the values of the constants up there.
First, we isolate x or y in a equation, let's do it with x in equation 1:
A*x = C - B*y
x = (C - B*y)/A
Now we can replace this in the equation 2, and solve it for y:
a*x + b*y = c
a*(C - B*y)/A + b*y = c
y*(b - a*B/A) = c - a*C/A
y = (c - a*C/A)/(b - a*B/A)
Now you have a solution for y, and we you know the value of y, you can put it in the eqution:
x = (C - B*y)/A
and find the value of x
Using brute force method to obtain the solution for the two system equations, from integers between -10 to 10 has only one solution at x = 3 and y = 2.
The given equation;
The solution of the given equation using brute force method is calculated as follows;
Let x value = 10
Let y value = -10
8(10) + 7(-10) = 10
3(10) - 5(-10) = 80
Let the value of x = -5
Let the value of y = 5
8(-5) + 7(5) = -5
3(-5) -5(5) = -40
Let the value of x = 0
Let the value of y = 0
8(0) + 7(0) = 0
3(0) - 5(0) = 0
Let the value of x = 2
Let the value of y = 1
8(2) + 7(1) = 23
3(2) -5(1) = 1
Let the value of x = 3
Let the value of y = 2
8(3) + 7(2) = 38
3(3) - 5(2) = -1
Thus, using brute force method to obtain the solution for the two system equation, from integers between -10 to 10 has only one solution at x = 3 and y = 2.
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