My teacher said we are supossed to solve this using systems of equations:
Let a, b, and c be real numbers such that a-7b+8c=4 and 8a+4b-c=7. Then a^2-b^2+c^2= ?
can someone please help?! i don't know what method to use to solve this or where to start
See answer here:
http://www.jiskha.com/display.cgi?id=1252885572
To solve this problem using systems of equations, you need to do the following steps:
Step 1: Write down the two given equations:
a - 7b + 8c = 4 (Equation 1)
8a + 4b - c = 7 (Equation 2)
Step 2: Choose a method to eliminate one variable. In this case, we will use the method of elimination.
Step 3: Multiply Equation 1 by 8 and Equation 2 by -1 to make the coefficients of 'a' the same:
8(a - 7b + 8c) = 8(4)
-1(8a + 4b - c) = -1(7)
This simplifies to:
8a - 56b + 64c = 32 (Equation 3)
-8a - 4b + c = -7 (Equation 4)
Step 4: Add Equation 3 and Equation 4 together to eliminate 'a':
(8a - 56b + 64c) + (-8a - 4b + c) = 32 + (-7)
This simplifies to:
-60b + 65c = 25 (Equation 5)
Step 5: Solve Equation 5 for one variable. Let's solve for 'b':
-60b + 65c = 25
-60b = -65c + 25
b = (-65c + 25)/(-60)
b = (65c - 25)/60 (Equation 6)
Step 6: Substitute the value of 'b' (from Equation 6) into any of the given equations. Let's substitute into Equation 1:
a - 7((65c - 25)/60) + 8c = 4
Multiply through by the denominator to get rid of the fraction:
60a - 7(65c - 25) + 480c = 240
Distribute and simplify:
60a - 455c + 175 + 480c = 240
Combine like terms:
60a + 25c = 65 (Equation 7)
Step 7: Solve Equation 7 for one variable. Let's solve for 'a':
60a + 25c = 65
60a = -25c + 65
a = (-25c + 65)/60 (Equation 8)
Step 8: Now, substitute the values of 'a' (from Equation 8) and 'b' (from Equation 6) into the expression a^2 - b^2 + c^2:
(((-25c + 65)/60)^2) - (((65c - 25)/60)^2) + c^2
Simplify this expression to find the value of a^2 - b^2 + c^2.