There's more than one way to combine them really
but an obvious one will be
![\bf \begin{array}{llll} h(x)&=&(f\circ g)(x)\\\\ &&\sqrt[3]{7x+1}\\\\ &&\sqrt[3]{g(x)}\leftarrow \begin{array}{llll} f(x)=\sqrt[3]{x }\\\\ g(x)=7x+1 \end{array} \end{array}\\\\ -----------------------------\\\\ (f\circ g)(x)\iff f[\quad g(x)\quad ]=\sqrt[3]{g(x)}\implies f[\quad g(x)\quad ]=\sqrt[3]{7x+1}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bllll%7D%0Ah%28x%29%26%3D%26%28f%5Ccirc%20g%29%28x%29%5C%5C%5C%5C%0A%26%26%5Csqrt%5B3%5D%7B7x%2B1%7D%5C%5C%5C%5C%0A%26%26%5Csqrt%5B3%5D%7Bg%28x%29%7D%5Cleftarrow%20%0A%5Cbegin%7Barray%7D%7Bllll%7D%0Af%28x%29%3D%5Csqrt%5B3%5D%7Bx%20%7D%5C%5C%5C%5C%0Ag%28x%29%3D7x%2B1%0A%5Cend%7Barray%7D%0A%5Cend%7Barray%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%28f%5Ccirc%20g%29%28x%29%5Ciff%20f%5B%5Cquad%20g%28x%29%5Cquad%20%5D%3D%5Csqrt%5B3%5D%7Bg%28x%29%7D%5Cimplies%20%20f%5B%5Cquad%20g%28x%29%5Cquad%20%5D%3D%5Csqrt%5B3%5D%7B7x%2B1%7D)
but an obvious one will be
Gwendolyn is asked to solve the equation -4(x-3) = 20. She shows the work below and states that the solution is x=-8
1 answer:
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We can start from the given line's coefficients and translate the line from the origin to the given point.
4(x -(-2)) -(y -3) = 0
4x +8 -y +3 = 0
The equation of the desired line is ...
4x -y = -11
_____
For standard form line ax+by=c, any parallel line will have only a different value of c. For c=0, the line goes through the origin (0, 0). To make it go through point (h, k) we can write it as
a(x-h) +b(y-k) = 0
which is completely equivalent to
ax +by = ah +bk
4(x -(-2)) -(y -3) = 0
4x +8 -y +3 = 0
The equation of the desired line is ...
4x -y = -11
_____
For standard form line ax+by=c, any parallel line will have only a different value of c. For c=0, the line goes through the origin (0, 0). To make it go through point (h, k) we can write it as
a(x-h) +b(y-k) = 0
which is completely equivalent to
ax +by = ah +bk