©
:
Pre

Calculus

Chapter 7A
Chapter 7A

Systems of Linear Equations
Geometry of Solutions
In an earlier chapter we learned how to solve a single equation in one unknown. The general form of
such an equation has the form
ax
b
,
where the constants
a
≠
0 and
b
are assumed known, and we are looking for a value of
x
which satisfies
the equation. Since
a
≠
0, the equation is easy to solve. Multiply by
a
−
1
. Thus, we have the solution
x
b
a
. The situation is not quite so simple when we have more than one equation and unknown. First
we give the standard form for a system of two equations in two unknowns.
ax
by
c
dx
ey
f
,
where the constants
a
,
b
,
c
,
d
,
e
, and
f
are assumed known.
By a solution of this system we mean a pair of numbers
x
0
and
y
0
which satisfy the system of
equations. That is, when the substitutions
x
x
0
and
y
y
0
are made in the system both equations
become identities.
This pair of numbers is commonly written as
x
0
,
y
0
and interpreted as a point in the Euclidean plane,
R
2
. By the solution set of a system we mean the totality of all possible solutions to the system.
Example 1
:
Which of the pairs of numbers
1,2
,
11,3
, or
−
19,
−
9
are solutions to the given
system?
2
x
−
5
y
7
−
x
2
y
1.
Solution:
To check that a pair of numbers is a solution we substitute the values in for
x
and
y
. In
the first equation if we substitute
x
1 and
y
2 we have
2
x
−
5
y
7
→
2
1
−
5
2
7
→
−
8
7
Since
−
8
≠
7 the first pair does not satisfy the first equation let alone both equations.
We try the second pair
11,3
next:
2
x
−
5
y
7
→
2
11
−
5
3
7
→
7
7
Okay, the second pair satisfies the first equation, but that is not enough to guarantee that we have a
solution to the system. The second equation still has to be checked, which we do below.
−
x
2
y
1
→
−
11
2
3
1
→
−
5
1
Since
−
5
≠
1, this pair does not solve the system.
©
:
Pre

Calculus
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