Question

Q#4 a. Draw the graph of linear function = 5 − 10 for values from −3 to 5. solve

(step by step with explanation and graph I give u 10 number)​

Answer 1

Step-by-step explanation:

Systems of Linear Equations

Often it is necessary to look at several functions of the same independent variable.  Consider the prior example where x, the number of items produced and sold, was the independent variable in three functions, the cost function, the revenue function, and the profit function.

            In general there may be:

                        n equations

                        v variables

Solving systems of equations

            There are four methods for solving systems of linear equations:

                        a.  graphical solution

                        b.  algebraic solution

                        c.  elimination method

                        d.  substitution method

Graphical solution

Example 1

            given are the two following linear equations:

                        f(x)  =  y  = 1 +  .5x

                        f(x)  =  y  = 11 -  2x

Graph the first equation by finding two data points.  By setting first x and then y equal to zero it is possible to find the y intercept on the vertical axis and the x intercept on the horizontal axis.

            If x = 0, then  f(0)  =  1  + .5(0)  =  1

            If y = 0, then  f(x)  =  0  = 1  +  .5x

                                                -.5x  =  1

                                                     x  =  -2

            The resulting data points are  (0,1)  and  (-2,0)

Graph the second  equation by finding two data points.  By setting first x and then y equal to zero it is possible to find the y intercept on the vertical axis and the x intercept on the horizontal axis.

            If x = 0, then  f(0)  =  11  - 2(0)  =  11

            If y = 0, then  f(x)  =  0  = 11  -  2x

                                                2x  =  11

                                                     x  =  5.5

            The resulting data points are  (0,11)  and  (5.5,0)

At the point of intersection of the two equations x and y have the same values.  From the graph these values can be read as x = 4 and y = 3.



 

Example 2

            given are the two following linear equations:

                        f(x)  =  y  = 15 -  5x

                        f(x)  =  y  = 25 -  5x

Graph the first equation by finding two data points.  By setting first x and then y equal to zero it is possible to find the y intercept on the vertical axis and the x intercept on the horizontal axis.

            If x = 0, then  f(0)  =  15  - 5(0)  =  15

            If y = 0, then  f(x)  =  0  = 15  -  5x

                                                5x  =  15

                                                     x  =  3

            The resulting data points are  (0,15)  and  (3,0)

Graph the second equation by finding two data points.  By setting first x and then y equal to zero it is possible to find the y intercept on the vertical axis and the x intercept on the horizontal axis.

            If x = 0, then  f(0)  =  25  - 5(0)  =  25

            If y = 0, then  f(x)  =  0  = 25  -  5x

                                                5x  =  25

                                                     x  =  5

            The resulting data points are  (0,25)  and  (5,0)

From the graph it can be seen that these lines do not intersect.  They are parallel.  They have the same slope.  There is no unique solution.



 

Example 3

            given are the two following linear equations:

                        21x - 7y  =  14

                        -15x  +  5y  =  -10

            Rewrite the equations by putting them into slope intercept form.

            The first equation becomes

                        7y  =  -14  +  21x

                          y  =  -2  +  3x

            The second equation becomes

                        5y  =  -10  +  15x

                          y  =  -2  +  3x

Graph either  equation by finding two data points.  By setting first x and then y equal to zero it is possible to find the y intercept on the vertical axis and the x intercept on the horizontal axis.

            If x = 0, then  f(0)  =  -2  +3(0)  =  -2

            If y = 0, then  f(x)  =  0  = -2  + 3x

                                                3x  =  2

                                                     x  =  2/3

            The resulting data points are  (0,-2)  and  (2/3,0)

From the graph it can be seen that these equations are equivalent.  There are an infinite number of solutions.



 

Algebraic solution

This method will be illustrated using supply and demand analysis.  This type of analysis is derived from the work of the great English economist Alfred Marshall.

Q  = quantity     and  P  =  price

P (s)= the supply function    and P (d)  = the demand function

When graphing price is placed on the vertical axis.  Thus price is the dependent variable.  It might be more logical to think of quantity as the dependent variable and this was the approach used by the great French economist, Leon Walras.  However by convention economists continue to graph using Marshall’s analysis which is referred to as the Marshallian cross.

The objective is to find an equilibrium price and quantity, i.e. a solution where price and quantity will have the same values in both the supply function and the price function.

            QE  = the equilibrium quantity           PE  = the equilibrium price

            For equilibrium 

                        supply  =  demand 

                        or P (s) = P (d)

Given the following functions

            P (s) =  3Q + 10  and            P (d) =  -1/2Q + 80

Set the equations equal to each other and solve for Q.

            P (s)  =  3Q + 10  =   -1/2Q + 80  = P (d)

                        3.5Q  =  70

                            Q  =  20                  The equilibrium quantity is 20.

Substitute this value for Q in either equation and solve for P.

            P (s)  =  3(20) + 10

            P (s) =  70

            P (d)  =  -1/2(20) + 80

            P (d)  =  70                            

Answer 2

$\huge{\underline{\bf\red{QuestiØn} :}}$

$\sf \red{Q4.} \ \ \ \ Draw \: the \: graph \: of \: linear \: function \\ \sf f(x) = 5x - 10 \: for \: values \: from \: - 3 \: to \: 5.$

$\huge{\underline{\bf\blue{SolutiØn}\ :}}$

On putting respective values of x in f(x) [ i.e.: y ] , we get the following results :—

• $\begin{aligned} \sf when \: x &= - 3, \\ \sf then, y & = 5( - 3) - 10 = - 25 \end{aligned}$

• $\begin{aligned} \sf when \: x &= - 2, \\ \sf then, y & = 5( - 2) - 10 = - 20 \end{aligned}$

• $\begin{aligned} \sf when \: x &= - 1, \\ \sf then, y & = 5( - 1) - 10 = - 15 \end{aligned}$

• $\begin{aligned} \sf when \: x &= 0, \\ \sf then, y & = 5(0) - 10 = - 10 \end{aligned}$

• $\begin{aligned} \sf when \: x &= 1, \\ \sf then, y & = 5(1) - 10 = - 5 \end{aligned}$

• $\begin{aligned} \sf when \: x &= 2, \\ \sf then, y & = 5(2) - 10 = 0 \end{aligned}$

• $\begin{aligned} \sf when \: x &= 3, \\ \sf then, y & = 5(3) - 10 = 5 \end{aligned}$

• $\begin{aligned} \sf when \: x &= 4, \\ \sf then, y & = 5(4) - 10 = 10 \end{aligned}$

• $\begin{aligned} \sf when \: x &= 5, \\ \sf then, y & = 5(5) - 10 = 15 \end{aligned}$

NOTE :

SCALE of the Graph :-

• On x - axis , 1 small division = 1 unit

• On y - axis , 1 small division = 5 units

So, let us point the following coordinates in the graph of the function f(x) :-

• A ( -3 , -25 )

• B ( -2 , -20 )

• C ( - 1 , - 15 )

• D ( 0 , - 10 )

• E ( 1 , - 5 )

• F ( 2 , 0 )

• G ( 3 , 5 )

• H ( 4 , 10 )

• I ( 5 , 15 )

And then, join the coordinates with ruler.

For graph ,

$[ \boxed{ \sf \ refer \ to \ the \ attachment \ }]$

$\red{\rule{5.5cm}{0.02cm}}$

$\Large{ \mid {\underline{\underline{\bf\green{BrainLiest \ AnswEr}}}} \mid }$