4x + 2 - 3x + 5 - 2(x + 5) = (2x - 5) + (3x + 4)
4x + 2 - 3x + 5 - 2x - 10 = 2x - 5 + 3x + 4
4x - 3x - 2x + 5 + 2 - 10 = 2x + 3x - 5 + 4
- x - 3 = 5x - 1
-x - 3 = 5x - 1 is the line above recopied
+x = +x
-3 = 6x - 1
<span> +1 = + 1 </span>
- 2 = 6x
- 2/6 = 6x/6
- 1/3 = x, the answer
Answer:
A. Z = 2 + N13
Step-by-step explanation:
Let's solve your equation step-by-step.
0=z2+4z−9
Step 1: Subtract z^2+4z-9 from both sides.
0−(z2+4z−9)=z2+4z−9−(z2+4z−9)
−z2−4z+9=0
For this equation: a=-1, b=-4, c=9
−1z2+−4z+9=0
Step 2: Use quadratic formula with a=-1, b=-4, c=9.
z=
−b±√b2−4ac
2a
z=
−(−4)±√(−4)2−4(−1)(9)
2(−1)
z=
4±√52
−2
z=−2−√13 or z=−2+√13
Answer:
Step-by-step explanation:
Okay, so I think I know what the equations are, but I might have misinterpreted them because of the syntax- I think when you ask a question you can use the symbols tool to input it in a more clear way, otherwise you can use parentheses and such.
Problem 1:
(x²)/4 +y²= 1
y= x+1
*substitute for y*
Now we have a one-variable equation we can solve-
x²/4 + (x+1)² = 1
x²/4 + (x+1)(x+1)= 1
x²/4 + x²+2x+1= 1
*subtract 1 from both sides to set equal to 0*
x²/4 +x^2+2x=0
x²/4 can also be 1/4 * x²
1/4 * x² +1*x² +2x = 0
*combine like terms*
5/4 * x^2+2x+ 0 =0
now, you can use the quadratic equation to solve for x
a= 5/4
b= 2
c=0
the syntax on this will be rough, but I'll do my best...
x= (-b ± √(b²-4ac))/(2a)
x= (-2 ±√(2²-4*(5/4)*(0))/(2*(5/4))
x= (-2 ±√(4-0))/(2.5)
x= (-2±2)/2.5
x will have 2 answers because of ±
x= 0 or x= 1.6
now plug that back into one of the equations and solve.
y= 0+1 = 1
y= 1.6+1= 2.6
Hopefully this explanation was enough to help you solve problem 2.
Problem 2:
x² + y² -16y +39= 0
y²- x² -9= 0
Answer:i guess yes
Step-by-step explanation:
Answer:
10!
Step-by-step explanation:
Because I ate chicken nuggets!