IXL - Approximate solutions using a table (Algebra 2 practice)

Use the table to estimate one solution, to the nearest tenth, of this equation:

–

The values in the table have been rounded to the nearest tenth.

x	– 2 x + 6	3 x
0	6	1
1	4	3
2	2	9
1.1	3.8	3.3
1.2	3.6	3.7
1.15	3.7	3.5

≈

ref_doc_title.

Learn with an example

question

Use the table to estimate one solution, to the nearest tenth, of this equation:

The values in the table have been rounded to the nearest tenth.

x	7 x	5 x + 12
0	1	12
1	7	17
2	49	22
1.5	18.5	19.5
1.6	22.5	20
1.55	20.4	19.8

≈

key idea

Any value of

that makes

(

)

and

(

)

equal is a solution of the equation

(

)

(

)

One way to find a solution of

(

)

(

)

is to try several

-values

that make the difference between

(

)

and

(

)

smaller and smaller. This method is called using successive

approximations.

solution

Look at the equation

You can think of the left side as the function

(

)

Similarly, you can think of the right side as the function

(

)

Now look at the table. The second and third columns give you values for the functions

(

)

and

(

)

Notice that when

1.5

(

)

is less than

(

)

but when

1.6

(

)

is greater than

(

)

x	f ( x ) = 7 x	g ( x ) = 5 x + 12
0	1	12
1	7	17
2	49	22
1.5	18.5	19.5
1.6	22.5	20
1.55	20.4	19.8

This means that there is an

-value

between

1.5

and

1.6

that makes

(

)

equal to

(

)

In the last row, when

1.55

(

)

is greater than

(

)

Since

(

)

is less than

(

)

when

1.5

but greater than

(

)

when

1.55

the solution of

is between

1.5

and

1.55

Values between 1.5 and 1.55 round down to 1.5. So,

≈

1.5

is one solution, to the nearest tenth, of

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