Answer:
169146
Step-by-step explanation:
Answer:
Therefore the solutions are
![x=0\\and\\ y =5\\\\Or\\\\x=-4\\and\\y=-3](https://tex.z-dn.net/?f=x%3D0%5C%5Cand%5C%5C%20y%20%3D5%5C%5C%5C%5COr%5C%5C%5C%5Cx%3D-4%5C%5Cand%5C%5Cy%3D-3)
Step-by-step explanation:
Given:
.........( 1 )
................( 2 )
To Find:
x = ?
y = ?
Solution:
Substituting ' y ' in Equation 1 we get
![x^{2}+(2x+5)^{2} =25](https://tex.z-dn.net/?f=x%5E%7B2%7D%2B%282x%2B5%29%5E%7B2%7D%20%3D25)
Using identity (A+B)²=A²+2AB+B² we get
![x^{2}+4x^{2}+20x+25=25\\\\5x^{2}+20x=0\\5x(x+4)=0\\5x=0\ or\ x+4=0\\x=0\ or\ x= -4](https://tex.z-dn.net/?f=x%5E%7B2%7D%2B4x%5E%7B2%7D%2B20x%2B25%3D25%5C%5C%5C%5C5x%5E%7B2%7D%2B20x%3D0%5C%5C5x%28x%2B4%29%3D0%5C%5C5x%3D0%5C%20or%5C%20x%2B4%3D0%5C%5Cx%3D0%5C%20or%5C%20x%3D%20-4)
Now Substitute x =0 in equation 2 we get
![y=2\times 0+5=5](https://tex.z-dn.net/?f=y%3D2%5Ctimes%200%2B5%3D5)
Or
Now Substitute x =-4 in equation 2 we get
![y=2\times -4+5=-3](https://tex.z-dn.net/?f=y%3D2%5Ctimes%20-4%2B5%3D-3)
Therefore the solutions are
![x=0\\and\\ y =5\\\\Or\\\\x=-4\\and\\y=-3](https://tex.z-dn.net/?f=x%3D0%5C%5Cand%5C%5C%20y%20%3D5%5C%5C%5C%5COr%5C%5C%5C%5Cx%3D-4%5C%5Cand%5C%5Cy%3D-3)
Answer:
Let x = the number
A negative number is 42 less than its square ⇒
x = x2 - 42
x2 - x - 42 = 0
(x - 7)(x + 6) = 0
x = 7 or -6
But since it is given that x is negative, the final answer is x = -6.
If this is the type of problem I’m thinking of it would be 26
Again same here, isolate the y variable to get it in y=mx+b form
6y=2x+15 divide by 6 to isolate the y and your equation is...
y=1/3x+15/6. Or 15/6 can be simplified to 5/2.