First Let we solve the Original system of equations:
equation (1): 
equation (2): 
Multiplying equation (1) by 7, we get
Subtracting,
implies 
Then
Thus the solution of the original equation is
Now Let we form the new equation:
Equation 2 is kept unchanged:
Equation (2):
Equation 1 is replaced with the sum of equation 1 and a multiple of equation 2:
Equation (1): 
Now solve this two equations: 
Multiply (1) by 7 and (2) by 8,
Subtracting,
implies
Then x=2.
so the solution for the new system of equation is x=2, y=1.
This Show that the solution to the system of equations 8x − 5y = 11 and 7x − 8y = 6 is the same as the solution to the given system of equations
You can write three equations in the numbers of nickels (n), dime (d), and quarters (q).
n + d + q = 23 . . . . . . . there are 23 coins total
0n +d -q = 2 . . . . . . . . .there are 2 more dimes than quarters
5n +10d +25q = 250 . .the total value is $2.50
The collection includes 11 nickels, 7 dimes, and 5 quarters.
_____
I used the matrix function of my calculator to solve these equations. You can find q by subtracting from the last equation five times the sum of the first two equations.
(5n +10d +25q) -5((n +d +q) +(d -q)) = (250) -5(23 +2)
25q = 125 . . . . . . . simplify
q = 5
From the second equation,
d = q +2 = 7
And from the first,
n = 23 -5 -7 = 11
Answer:
4 √6
Step-by-step explanation:
We have a few right triangles. We know that a²+b²=c², with c being the side opposite the right angle. Representing the side without a value as z, we have:
m²+z² = (8+4)² = 12²
4²+n²=z²
8²+n²=m²
We have 3 equations with 3 unknown variables, so this should be solvable. One way to find a solution is to put everything in terms of m and go from there. First, we can take n out of the equations entirely, removing one variable. We can do this by solving for it in terms of z and plugging that into the third equation, removing a variable as well as an equation.
4²+n²=z²
subtract 4²=16 from both sides
z²-16 = n²
plug that into the third equation
64 + z² - 16 = m²
48 + z² = m²
subtract 48 from both sides to solve for z²
z² = m² - 48
plug that into the first equation
m² + m² - 48 = 144
2m² - 48 = 144
add 48 to both sides to isolate the m² and its coefficient
192 = 2m²
divide both sides by 2 to isolate the m²
96 = m²
square root both sides to solve for m
√96 = m
we know that 96 = 16 * 6, and 16 = 4², so
m = √96 = √(4²*6) = 4 √6
Pop ok you see so the answer here would be frantically elastic
