1.73 is the answer >:-)............................
In each case, you can use the second equation to create an expression for y that will substitute into the first equation. Then you can write the result in standard form and use any of several means to find the number of solutions.
System A
x² + (-x/2)² = 17
x² = 17/(5/4) = 13.6
x = ±√13.6 . . . . 2 real solutions
System B
-6x +5 = x² -7x +10
x² -x +5 = 0
The discriminant is ...
D = (-1)²-4(1)(5) = -20 . . . . 0 real solutions
System C
y = 8x +17 = -2x² +9
2x² +8x +8 = 0
2(x+2)² = 0
x = -2 . . . . 1 real solution
Answer:
7
Step-by-step explanation:
Simplifying
5x(4y + 3x) = 5x(3x + 4y)
Reorder the terms:
5x(3x + 4y) = 5x(3x + 4y)
(3x * 5x + 4y * 5x) = 5x(3x + 4y)
Reorder the terms:
(20xy + 15x2) = 5x(3x + 4y)
(20xy + 15x2) = 5x(3x + 4y)
20xy + 15x2 = (3x * 5x + 4y * 5x)
Reorder the terms:
20xy + 15x2 = (20xy + 15x2)
20xy + 15x2 = (20xy + 15x2)
Add '-20xy' to each side of the equation.
20xy + -20xy + 15x2 = 20xy + -20xy + 15x2
Combine like terms: 20xy + -20xy = 0
0 + 15x2 = 20xy + -20xy + 15x2
15x2 = 20xy + -20xy + 15x2
Combine like terms: 20xy + -20xy = 0
15x2 = 0 + 15x2
15x2 = 15x2
Add '-15x2' to each side of the equation.
15x2 + -15x2 = 15x2 + -15x2
Combine like terms: 15x2 + -15x2 = 0
0 = 15x2 + -15x2
Combine like terms: 15x2 + -15x2 = 0
0 = 0
Solving
0 = 0
Couldn't find a variable to solve for.
This equation is an identity, all real numbers are solutions.
This can be rewritten as
(x -6)² = 90 . . . . . the left side is already a perfect square
x -6 = ±√90 . . . . . take the square root
x = 6 ±3√10 . . . . add 6
