Informative line

Writing Equations Using Tables Figures And Graphs

Drawing Table through Equation

  • We can draw a table through an equation involving two variables.
  • In an equation of two variables, one of the variables is input (independent) and other one is output (dependent).
  • To draw the table, give some values (number) to input to find the value of output.

For example: \(y=x-1\)

\(x=\) input/independent variable and \(y=\) output/dependent variable

For \(x=1\),

\(y=1-1\)

\(y=0\)

Thus, when \(x=1,\;y=0\)

For \(x=2\),

\(y=2-1\)

\(y=1\)

Thus, when \(x=2,\;y=1\)

For \(x=3\),

\(y=3-1\)

\(y=2\)

Thus, when \(x=3,\;y=2\)

Thus, table for the given equation is:

\(\begin{array} {|c|c|} \hline x(\text{input})&y(\text{output})\\ \hline 1&0\\ 2&1\\ 3&2\\ \hline \end{array}\)

  • The values of \(x\) and the corresponding values of \(y\) are called ordered pairs.

Thus, \((1,\;0),\;(2,\;1)\) and \((3,\;2)\) are ordered pairs of the equation: \(y=x-1\)

Illustration Questions

Which one of the following tables contains the values that satisfy the equation:   \(x=2y-1\)

A \(\begin{array} {|c|c|} \hline y(\text{input})&x(\text{output})\\ \hline 1&1\\ -1&3\\ \hline \end{array}\)

B \(\begin{array} {|c|c|} \hline y(\text{input})&x(\text{output})\\ \hline 1&1\\ 2&3\\ \hline \end{array}\)

C \(\begin{array} {|c|c|} \hline y(\text{input})&x(\text{output})\\ \hline -2&1\\ 0.5&0\\ \hline \end{array}\)

D \(\begin{array} {|c|c|} \hline y(\text{input})&x(\text{output})\\ \hline -0.5&-3\\ -3&-2.5\\ \hline \end{array}\)

×

Here, \(y=\) input/independent variable and \(x=\) output/dependent variable

Put the values of \(y\) as input and find the values of \(x\) as output.

In option (A),

For \(y=1,\)

\(x=2(1)-1\)

\(x=2-1\)

\(x=1\)

For \(y=-1\)

\(x=2(-1)-1\)

\(x=-2-1\)

\(x=-3\)

The given values are:

For \(y=1,\;x=1\) 

For \(y=-1,\;x=3\)

Hence, option (A) is incorrect.

In option (B),

For \(y=1\),

\(x=2(1)-1\)

\(x=2-1\)

\(x=1\)

For \(y=2\),

\(x=2(2)-1\)

\(x=4-1\)

\(x=3\)

The given values are:

For \(y=1,\;x=1\)

For \(y=2,\;x=3\)

Hence, option (B) is correct.

In option (C),

For \(y=-2\)

\(x=2(-2)-1\)

\(x=-4-1\)

\(x=-5\)

For \(y=0.5\),

\(x=2(0.5)-1\)

\(x=1.0-1\)

\(x=0\)

The given values are:

For \(y=-2,\;x=1\)

For \(y=0.5,\;x=0\)

Hence, option (C) is incorrect.

In option (D),

For \(y=-0.5\),

\(x=2(-0.5)-1\)

\(x=-1.0-1\)

\(x=-2.0\;\text{or}\;-2\)

For \(y=-3\),

\(x=2(-3)-1\)

\(x=-6-1\)

\(x=-7\)

The given values are:

For \(y=-0.5,\;x=-2\)

For \(y=-3,\;x=-2.5\)

Hence, option (D) is incorrect.

Which one of the following tables contains the values that satisfy the equation:   \(x=2y-1\)

A

\(\begin{array} {|c|c|} \hline y(\text{input})&x(\text{output})\\ \hline 1&1\\ -1&3\\ \hline \end{array}\)

.

B

\(\begin{array} {|c|c|} \hline y(\text{input})&x(\text{output})\\ \hline 1&1\\ 2&3\\ \hline \end{array}\)

C

\(\begin{array} {|c|c|} \hline y(\text{input})&x(\text{output})\\ \hline -2&1\\ 0.5&0\\ \hline \end{array}\)

D

\(\begin{array} {|c|c|} \hline y(\text{input})&x(\text{output})\\ \hline -0.5&-3\\ -3&-2.5\\ \hline \end{array}\)

Option B is Correct

Equations through Models

  • We can write algebraic equations through models.
  • Consider an example to understand it.

Write an algebraic equation for the following model.

\(\triangle\triangle\triangle\bigcirc\bigcirc=\bigcirc\bigcirc\bigcirc\bigcirc\triangle\)................(1)

  • To write an algebraic equation, follow the given steps:

Step 1: Allot different variables to different figures.

Let \(\triangle=x\;\;\;\bigcirc=y\)

Step 2: Write individual expressions for both the groups that are present on both the sides of the equals sign.

 

 

For      \(\triangle\triangle\triangle\bigcirc\bigcirc\)

It is the group of three triangles 'and' two circles.

Here, 'and' represents addition (keyword)

\(\therefore\) Expression for this :

\(3x+2y\)     \((\triangle=x,\;\;\;\bigcirc=y)\)

 

 

Similarly,         \(\bigcirc\bigcirc\bigcirc\bigcirc\triangle\)

It is the group four circles 'and' one triangle

Here, 'and' represents addition (keyword)

\(\therefore\) Expression for this:

\(4y+x\)        \((\triangle=x,\;\;\;\bigcirc=y)\)

Step 3: Now put the expressions of both the groups in place of model to get an algebraic equation.

Thus, algebraic equation for the given model is:

\(3x+2y=4y+x\)

Note: We can also write an algebraic equation through a weighing scale (balancing diagram) which has equal weight on both the sides.

Illustration Questions

Write an algebraic equation for the given balancing diagram. Each \(\triangle\) weighs \(x\) pound and each \(\bigcirc\) weighs \(y\) pound.

A \(3x+4y=6x+3y\)

B \(3x+4y=2x+6y\)

C \(7x=8y\)

D \(5x=9y\)

×

Given: \(\triangle=x\) pound and \(\bigcirc=y\) pound

Since, both the sides of the balancing diagram are balanced.

image

Write individual expressions for both the groups that are present on both the sides of the equals sign.

For

image

This is a group of three \(\triangle\)'s 'and' four \(\bigcirc\)'s

Here, 'and' represents addition (keyword)

\(\therefore\) Expression for this:

\(3\triangle+4\bigcirc\)

or

\(3x+4y\)         \((\because\;\triangle=x,\;\;\;\bigcirc=y)\)

For

image

This is a group of two \(\triangle\)'s 'and' six \(\bigcirc\)'s

Here, 'and' represents addition (keyword)

\(\therefore\) Expression for this:

\(2\triangle+6\bigcirc\)

or

\(2x+6y\)        \((\because\;\triangle=x,\;\;\;\bigcirc=y)\)

Now put the expressions of both the groups in place of model to get an algebraic equation.

Thus, algebraic equation is:

\(3x+4y=2x+6y\)

Hence, option (B) is correct.

Write an algebraic equation for the given balancing diagram. Each \(\triangle\) weighs \(x\) pound and each \(\bigcirc\) weighs \(y\) pound.

image
A

\(3x+4y=6x+3y\)

.

B

\(3x+4y=2x+6y\)

C

\(7x=8y\)

D

\(5x=9y\)

Option B is Correct

Writing Equation through Table

  • When values of variables are given through a table as input and output values, we can write an algebraic equation from the table.
  • We can write an equation through the given input-output table using hit and trial method to check how each input is related to corresponding output.
  • We would guess the relation between input and output using single or multiple operations and then apply them to check.

For example:

\(\begin{array} {|c|c|} \hline x(\text{input})&y(\text{output})\\ \hline 3&12\\ 4&16\\ 5&20\\ \hline \end{array}\)

\(\to\) We can see that all the values of \(y\) are greater than all the corresponding values of \(x.\)

\(\to\) Therefore, addition or multiplication operation must be performed.

\(\to\) Let's check for addition.

\(12\) can be obtained by adding \(3\) to \(9,\) i.e.

\(3+9=12\)

Check for other values of \(y\).

\(4+9=13,\;\;5+9=14\)

Thus, the addition operation is not working here.

\(\to\) Now let's check for multiplication.

\(12\) can be obtained by multiplying \(3\) by \(4,\) i.e.

\(3×4=12\)

Check for other values of \(y\)

\(4×4=16\\ 5×4=20\)

\(\to\) Thus, the multiplication by \(4\) satisfies the relation between input and output values of the given table.

\(\therefore\) Algebraic equation for given table is: \(y=4x\)

Note:

  1. When the values of output are less than all the corresponding values of input then subtraction or division must be performed there but other operations can be used as well.
  2. When the values of output are greater than all the corresponding values of input then addition or multiplication must be performed there but other operations can be used as well.

Illustration Questions

Write an algebraic equation for the following input-output table.   \(\begin{array} {|c|c|} \hline x(\text{input})&y(\text{output})\\ \hline 16&9\\ 14&8\\ 12&7\\ \hline \end{array}\)

A \(y=\dfrac{x}{2}\)

B \(y=\dfrac{x}{2}+1\)

C \(y=\dfrac{x}{2}-1\)

D \(x=\dfrac{y}{2}+1\)

×

We can observe that all the values of \(y\) are less than all the corresponding values of \(x.\)

\(\therefore\) Subtraction or division operation must be performed first.

Let's check for the subtraction.

\(9\) can be obtained by subtracting \(7\) from \(16\), i.e.

\(16-7=9\)

Now check for other values of \(y\).

\(14-7=7\\ 12-7=5\)

Thus, the subtraction operation is not working here.

Now let's check for the division.

\(9\) can not be obtained by dividing \(16\) by any number.

But \(9\) can be obtained by dividing \(16\) by \(2\) and then adding \(1,\) i.e.

\((16\div2)+1=8+1=9\)

Let's check for other values of \(y\).

\((14\div2)+1=7+1=8\\ (12\div2)+1=6+1=7\)

Thus, the division by \(2\) and the addition of \(1,\) satisfies the relation between input and output values of the given table.

\(\therefore\) Algebraic equation for the given table is:

\(y=\dfrac{x}{2}+1\)

Hence, option (B) is correct.

Write an algebraic equation for the following input-output table.   \(\begin{array} {|c|c|} \hline x(\text{input})&y(\text{output})\\ \hline 16&9\\ 14&8\\ 12&7\\ \hline \end{array}\)

A

\(y=\dfrac{x}{2}\)

.

B

\(y=\dfrac{x}{2}+1\)

C

\(y=\dfrac{x}{2}-1\)

D

\(x=\dfrac{y}{2}+1\)

Option B is Correct

Dependent and Independent Variables

Definition: 

  • An independent variable is a variable with a subject of choice, whose values can be changed. Its value is not affected by the values of another variable.
  • The value of a dependent variable depends upon the value of an independent variable. 
  • To identify a variable to be dependent or independent, fetch the important information from the given problem.

For example:

Alex earns \($10\) per hour for trimming lawns.

  • The amount of money he earns, \(m\) , depends upon how many hours he works for, \(h\).
  • The more hours he works for, the more money he earns. Therefore, the dependent variable is money, \(m\), and the independent variable is hours, \(h\).
  • The number of hours he works for does not depend upon the money he earns.

So, the independent variable \(=h\)

and the dependent variable \(=m\)

Illustration Questions

Tina is plucking flowers to make a bouquet. The more flowers she plucks, the bigger bouquet she will be able to make. Which one of following options represents the dependent variable?

A Number of flowers

B Size of bouquet

C Both (A) and (B)

D None of these

×

The size of the bouquet depends upon the number of flowers.

The more flowers she plucks, the bigger bouquet she will be able to make.

Thus, the dependent variable is the size of the bouquet.

Hence, option (B) is correct.

Tina is plucking flowers to make a bouquet. The more flowers she plucks, the bigger bouquet she will be able to make. Which one of following options represents the dependent variable?

A

Number of flowers

.

B

Size of bouquet

C

Both (A) and (B)

D

None of these

Option B is Correct

Equations with Graphs

A graph shows the relationship between two variables.

When graphing:

(i) The independent variable is always across the \(x-\)axis.

(ii) The dependent variable is always up the \(y-\)axis.

Coordinates: A pair of values that shows an exact position on a coordinate plane.

For example: A bakery shopkeeper makes cookies. The table shows the relationship between the time in hours \((x)\) and the number of cookies the shopkeeper makes \((y)\).

  • We can write an equation from the given table.

\(\begin{array} {|c|c|} \hline (x)&(y)\\ \text{(In hours)}&\text{(Number of cookies)}\\ \hline 1&20\\ \hline 2&40\\ \hline 3&60\\ \hline 4&80\\ \hline \end{array}\)

  • The equation for the situation is, \(y=20x\)
  • Thus, \(20\) cookies are being made each hour.

  • The value of \(y\) depends upon the value of \(x.\)

Hence, \(y\) (the number of cookies) is the dependent variable.

\(x\) (the time in hours) is the independent variable.

  • Now, write coordinates from the given table.

\(\begin{array} {|c|c|} \hline (x)&(y)&\text{Coordinates}\\ &&(x,\,y)\\ \hline 1&20&(1,\;20)\\ \hline 2&40&(2,\;40)\\ \hline 3&60&(3,\;60)\\ \hline 4&80&(4,\;80)\\ \hline \end{array}\)

  • Plot the coordinates on the graph from the table.
  • From this plotted graph, we can observe the relation between \(x\) and \(y.\)
  • If one value increases, the other value also increases.

To write an equation from a given graph:

Try to observe the relation between the coordinate pairs and write the relation in an equation form.

For example:

The following graph shows the relationship between the number of miles traveled and the number of hours.

  • Write coordinates \((x,\;y)\) from the given graph:

\((2,\;10)\\ (4,\;20)\\ (6,\;30)\)

  • Try to find the relation between the pairs.

The first coordinate is, \((2,\;10)\)

Here, \(2\) is multiplied with \(5\) to get \(10\).

The second coordinate is, \((4,\;20)\)

\(4\) is multiplied with \(5\) to get \(20\).

The third coordinate is \((6,\;30)\)

\(6\) is multiplied with \(5\) to get \(30\).

So, we can write a rule: \(y=5x\)

  • We will check to be sure that our rule is working or not.

\(y=5x\)

Put \(x=1\), then \(y=5×1=5\)

Put \(x=2\), then  \(y=5×2=10\)

Put \(x=3\), then \(y=5×3=15\) and so on.

It means our rule is working.

  • So, \(y=5x\) is the required equation.
  • After finding the rule for the graph, we can get next coordinates.
  • We can use trial and error method also, to find the rule for the graph from the given options.

Illustration Questions

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A

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B

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C

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D

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×

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Option B is Correct

Practice Now