$0.03+$0.04=$0.15x4=$0.60
I'm going to use the substitution method.
If y = - 3/2 - 7, then:
1/2x + 5 = - 3/2x - 7
Combine like terms:
4/2x = - 12. Mutiply both sides by 2/4 to get x = -12 (2/4)
Simplify to get:
x = - 6.
Plug - 6 back in for x in either equation.
Y = 1/2( - 6) + 5 which becomes Y = - 3 + 5.
X = - 6, Y = 2
Firstly start off with finding the lowest common multiple of 14 and 24. The LCM would be 168. Then you would look at your powers, here you have to choose the highest power of each letter. So 14x^4 and 24x^6 the highest is x^6, then for y it would be y^6 and then for z it would be z^8. So altogether your answer would be 168x^6y^6z^8
Y because if you add them you will get 10x =-10
The y will be eliminated because you have positive 3 -4
![\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \stackrel{\textit{we'll use this one}}{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{2}{ h},\stackrel{-1}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=2\\ k=-1 \end{cases}\implies y=a(x-2)^2-1 \\\\\\ \textit{we also know that } \begin{cases} y=0\\ x=5 \end{cases}\implies 0=a(5-2)^2-1\implies 1=9a \\\\\\ \cfrac{1}{9}=a\qquad therefore\qquad \boxed{y=\cfrac{1}{9}(x-2)^2-1}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Bparabola%20vertex%20form%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20%5Cstackrel%7B%5Ctextit%7Bwe%27ll%20use%20this%20one%7D%7D%7By%3Da%28x-%20h%29%5E2%2B%20k%7D%5C%5C%5C%5C%20x%3Da%28y-%20k%29%5E2%2B%20h%20%5Cend%7Barray%7D%20%5Cqquad%5Cqquad%20vertex~~%28%5Cstackrel%7B2%7D%7B%20h%7D%2C%5Cstackrel%7B-1%7D%7B%20k%7D%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20h%3D2%5C%5C%20k%3D-1%20%5Cend%7Bcases%7D%5Cimplies%20y%3Da%28x-2%29%5E2-1%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Bwe%20also%20know%20that%20%7D%20%5Cbegin%7Bcases%7D%20y%3D0%5C%5C%20x%3D5%20%5Cend%7Bcases%7D%5Cimplies%200%3Da%285-2%29%5E2-1%5Cimplies%201%3D9a%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B1%7D%7B9%7D%3Da%5Cqquad%20therefore%5Cqquad%20%5Cboxed%7By%3D%5Ccfrac%7B1%7D%7B9%7D%28x-2%29%5E2-1%7D)
now, let's expand the squared term to get the standard form of the quadratic.
