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Lynna [10]
3 years ago
14

Solve the following simultaneous equation using either the substitution method or the elimination method, showing all the necess

ary steps.
Will award brainliest!!

Mathematics
1 answer:
slavikrds [6]3 years ago
4 0

Answer:

x=1\frac{83}{91}\\y=-2\frac{73}{91}

Step-by-step explanation:

#Using the method of substitution and replacement of unknown values.

Given that

\frac{3}{2}x+\frac{2}{3}y=1...(i)\\and\\\frac{1}{6}x-\frac{3}{5}y=2...(ii)\\

Express eqtn (ii) in terms of x

\frac{1}{6}x=2+\frac{3}{5}y\\x=12+\frac{18}{5}y

Replacing for x in eqtn (i)

\frac{3}{2}(12+\frac{18}{5}y)+\frac{2}{3}y=1\\18+\frac{27}{5}y+\frac{2}{3}y=1\\18-1=-\frac{91}{15}y\\y=-2\frac{73}{91}

Substitute y value in eqtn(ii) to obtain x

Y=-2\frac{73}{91}\\\frac{1}{6}x-\frac{3}{5}y=2\\x=2+\frac{3}{5}(-2\frac{73}{91})\\x=1\frac{83}{91}

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Find f(a), f(a+h), and<br> 71. f(x) = 7x - 3<br> f(a+h)-f(a)<br> h<br> if h = 0.<br> 72. f(x) = 5x²
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71. \ \ \ f(a) \  = \  7a \ - \ 3; \ f(a+h) \  =  \ 7a \ + \ 7h \ - \ 3; \ \displaystyle\frac{f(a+h) \ - \ f(a)}{h} \ = \ 7

72. \ \ \ f(a) \  = \  5a^{2}; \ f(a+h) \  =  \ {5a}^{2} \ + \ 10ah \ + \ {5h}^{2}; \ \displaystyle\frac{f(a+h) \ - \ f(a)}{h} \ = \ 10a \ + \ 5h

Step-by-step explanation:

In single-variable calculus, the difference quotient is the expression

                                              \displaystyle\frac{f(x+h) \ - \ f(x)}{h},

which its name comes from the fact that it is the quotient of the difference of the evaluated values of the function by the difference of its corresponding input values (as shown in the figure below).

This expression looks similar to the method of evaluating the slope of a line. Indeed, the difference quotient provides the slope of a secant line (in blue) that passes through two coordinate points on a curve.

                                             m \ \ = \ \ \displaystyle\frac{\Delta y}{\Delta x} \ \ = \ \ \displaystyle\frac{rise}{run}.

Similarly, the difference quotient is a measure of the average rate of change of the function over an interval. When the limit of the difference quotient is taken as <em>h</em> approaches 0 gives the instantaneous rate of change (rate of change in an instant) or the derivative of the function.

Therefore,

              71. \ \ \ \ \ \displaystyle\frac{f(a \ + \ h) \ - \ f(a)}{h} \ \ = \ \ \displaystyle\frac{(7a \ + \ 7h \ - \ 3) \ - \ (7a \ - \ 3)}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{7h}{h} \\ \\ \-\hspace{4.25cm} = \ \ 7

               72. \ \ \ \ \ \displaystyle\frac{f(a \ + \ h) \ - \ f(a)}{h} \ \ = \ \ \displaystyle\frac{{5(a \ + \ h)}^{2} \ - \ {5(a)}^{2}}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{{5a}^{2} \ + \ 10ah \ + \ {5h}^{2} \ - \ {5a}^{2}}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{h(10a \ + \ 5h)}{h} \\ \\ \-\hspace{4.25cm} = \ \ 10a \ + \ 5h

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