For this type of problem you would have to move the decimal as many places to the right as needed to make the divisor a whole number the you would move the dividend the same amount of times even if it is a whole number . Then you use what you have to divide normally. Hope this helps!
Answer:
See below
Step-by-step explanation:
4(x + 5) = 9x + 4x − 34
4x + 20 = 9x + 4x − 34 [Distributive]
4x + 20 = 13x − 34 {Combine Like Terms]
4x - 13x + 20 = 13x - 13x − 34 [Subtractive: -13x both sides]
-5x + 20 = - 34 [Combine Like Terms]
-5x + 20 - 20 = -34 - 20 [Subtractive: - 20 both sides]
-5x = -54 [Combine Like Terms]
x = -54/-5 [Division Property: divide both sides by -5]
x = - 10 4/5
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
I like the cat on your pfp LOL sorry i couldn’t help tho
