Determine how much was invested at each rate.
Let x represent the amount invested at 4%
Let y represent the amount invested at 5%
Total amount invested is $5 000 so,
x + y = 5 000
Recall that
I = prt
or
interest = principal x rate x time
For
1
year:
Interest earned on
x
dollars invested at 4% is
0.04x
Interest earned on
y
dollars invested at 5% is
0.05y
Total interest earned is $230
so,
0.04x + 0.05y = 230
For this problem, the linear system is:
x + y = 5 000
0.04x + 0.05y = 230
This system can be solved by isolating
either
x or y in the
first
equation then
substituting
into the
second
equation.
On your own
, complete the solution to find that $2 000 will be invested at 4% and
$3 000 will be invested at 5%.
The correct solution is:
(2000,3000), or:
$2000.00 was invested at 4%, and $3000.00
was invested at 5%.
Time for an exercise!
In your textbook, read pages 417
–
424.
On pages 425
–
426, do
questions 4
–
22.
Example 2 on page 420 can also be viewed as an
animation in the e-book.
Look for
Lesson 7.4 Using a Substitution Strategy
to Solve a System of Linear Equations Example 2a
and
Example
2b
.

10C
–
C5-6
–
LG
–
Systems of Linear Equations
Page 19 of 42
Solving linear systems of equations by substitution may require us to multiply an
equation by a non-zero number in order to eliminate any fractional coefficients.
It is
important to note the following.
SOLVING LINEAR SYSTEMS OF EQUATIONS USING ELIMINATION
There is one final algebraic strategy for solving linear systems of equations that
we will study.
The
elimination strategy
requires us to
eliminate
one variable
(temporarily) in order to solve for the variable that remains.
We then substitute the
known value into one of the original equations to solve for the variable that was originally
eliminated.
When using the elimination strategy, the two equations are either
added
or
subtracted
in order to eliminate one variable.
Carefully study the next example.
Example
Solve the following system using the
elimination strategy
.
3x
–
5y = -9
4x + 5y = 23
Do you notice that the
coefficients
of
y
in the two equations are
opposites
or
additive
inverses
of one another?
Since opposites or additive inverses
add to 0
,
adding
the
two equations will
eliminate y
.
3x
–
5y = -9
4x + 5y = 23
7x + 0y = 14
7x = 14
one equation in one variable
x = 2
We now substitute x = 2 into
either
equation to solve for y.
Let‟s substitute into the
second equation because there are no negative signs!
Multiplying or dividing both sides of an equation by a non-zero number
does not change the solution but simply produces an equivalent equation.
Line up like terms
vertically then add!
y is eliminated!

10C
–
C5-6
–
LG
–
Systems of Linear Equations
Page 20 of 42
4x + 5y = 23
4(2) + 5y = 23
8 + 5y = 23
5y = 15
y = 3
The solution is
(2, 3)
.
Let‟s check by substituting the point into
both
equations:
Let‟s check by substituting the point into
both
equations:
3x
–
5y = -9
4x + 5y = 23
3(2)
–
5(3)
-9
4(2) + 5(3)
23
6
–
15
8 + 15
-9
=
-9
23
=
23
LS = RS
LS = RS
Therefore, (2, 3) is the solution.
You can also check by
graphing
but remember to first
rewrite each equation in the form
y = mx + b
.

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- Winter '19
- jane smith
- Math, Linear Equations, Equations, Systems Of Linear Equations, Elementary algebra