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# Solving Pairs of Equations Help (page 2)

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By — McGraw-Hill Professional
Updated on Oct 3, 2011

## Solving Pairs of Equations Practice Problems

#### PROBLEM 1

Using the Cartesian plane to plot their graphs, what can be said about the solutions to the simultaneous equations y = x + 3 and ( x − 1) 2 + ( y + 2) 2 = 9?

#### SOLUTION 1

The graphs of these equations are shown in Fig. 6-14. The equation y = x + 3 has a graph that is a straight line, ramping up toward the right with slope equal to 1, and intersecting the y axis at (0,3). The equation ( x − 1) 2 + ( y + 2) 2 = 9 has a graph that is a circle whose radius is 3 units, centered at the point (1, −2). It is apparent that this line and circle do not intersect. This means that there exist no solutions to this pair of simultaneous equations.

Fig. 6-14 . Illustration for Problem 1.

#### PROBLEM 2

Using the Cartesian plane to plot their graphs, what can be said about the solutions to the simultaneous equations y = 1 and ( x − 1) 2 + ( y + 2) 2 = 9?

#### SOLUTION 2

The graphs of these equations are shown in Fig. 6-15. The equation y = 1 has a graph that is a horizontal straight line intersecting the y axis at (0,1). The equation ( x − 1) 2 + ( y + 2) 2 = 9 has a graph that is a circle whose radius is 3 units, centered at the point (1, −2). It appears from the graph that the equations have a single common solution denoted by the point (1, 1), indicating that x = 1 and y = 1.

Fig. 6-15 . Illustration for Problem 2

Let’s use algebra to solve the equations and see if the graph tells us the true story. Substituting 1 for y in the equation of a circle (because one of the equations tells us that y = 1), we get a single equation in a single variable:

( x − 1) 2 + (1 + 2) 2 = 9

( x − 1) 2 + 3 2 = 9

( x − 1) 2 + 9 = 9

( x − 1) 2 = 0

x − 1 =0

x = 1

It checks out. There is only one solution to this pair of simultaneous equations, and that is x = 1 and y = 1, denoted by the point (1, 1).

Practice problems for these concepts can be found at: The Cartesian Plane Practice Test.

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