Practice Problems, Mat110, Chapter 9.
(1) Use the substitution method to solve the systems of linear equations:
(a) x - 4y = 9
2x + 3y = -4
(b) 3x + y = 1
2x - 4y =-32
(2) Use the elimination method to solve the system of linear equations:
(a) x + 2y = 6
2x - 3y = 5
(b) 2x - 5y = -4
3x + 7y = 23
(c)
x + 5y + z = 7
2x + 3y - 3z = -13
4x - y + 2z = -1
(d)
x + 2y - 3z = -7
2x – y + 4z = 11
4x + 3y – 4z = -3
(3) Use the substitution or (graphical) intersection method to solve the system of nonlinear equations:
 y  x2  2x  3
(a) 
 y  x 1
x  2 y  3
(b) 
 xy  1
2
2
 2 x  y  9
(c)  2
2
 x  y  3
(4) Decompose into partial fractions:
(a)
x
x  8 x  15
(b)
5
x  6x
2
2
Solutions:
(1)
The basic strategy for solving a system of 2 linear equations in 2 variables is:
(i) Solve one of the equations for one of the variables (tip: choose a variable with
coefficient 1 if you can, it’s usually easiest to deal with).
(ii) Substitute this solution into the other equation, which will give you an equation with a
single variable (which can be solved by isolation).
(iii) Now you need to go back and find the value of the other variable to make your
ordered pair – this is known as ‘back-solving'.
Alright, rock-action:
(a)
x - 4y = 9
2x + 3y = -4
x - 4y = 9
+4y +4y
x = 9 + 4y
2(9 + 4y) + 3y = -4
18 + 8y + 3y = -4
-18
-18
11y = -22,
y = -2.
(In this system, we solve for the x in the first equation,
since it has coefficient 1).
(now, we substitute this solution, 9 + 4y, in for x
in the second (other) equation.)
(now back-solve for the x - you can use the equation
x = 9 + 4y)
x = 9 + 4(-2) = 1.
The solution is the ordered pair (1,-2).
(b) 3x + y = 1
2x - 4y = -32
3x + y = 1
-3x
-3x
y = 1 - 3x,
2x - 4(1 - 3x) = -32,
2x - 4 + 12x = -32,
+4
+4
14x = -28,
x = -28/14,
x = -2.
(here, we begin by solving for y in the first equation, since
it's the only variable with coefficient 1).
(now, substitute 1 - 3x in for y in the other equation:)
(Isolate x)
(Now, back-solve for y to complete the ordered pair:)
y = 1 - 3(-2),
y = 7.
The solution is the ordered pair (-2, 7).
(2) The elimination method is based on the fact that, within a linear system, we can
multiply any equation by a constant, and add the result to the other equation(s), and the
new equations will have the same solutions as the original ones. Generally, when you
want to eliminate a variable, you look at the coefficients on the variable, figure out the
least number they both go into - that's the Least Common Multiple or LCM, and multiply
so that one of the coefficients becomes the LCM, and the other -LCM.
Let's rip:
(a) x + 2y = 6,
2x - 3y = 5
(I'm going to eliminate the x - note that 1 and 2 are the coefficients on
x, and LCM(1,2) = 2 - the coefficient on the second variable is
already +2, so let's make the second -2 by multiplying by -2:)
-2(x + 2y = 6)
2x - 3y = 5
-2x - 4y = -12
2x - 3y = 5
-7y = -7,
y = 1.
(Now use this value to back-solve for x - you can use either equation:)
x + 2(1) = 6
-2
-2
x = 4.
So, the solution is the ordered pair (4,1)
(b) 2x - 5y = -4
3x + 7y = 23
(This time I'm going to eliminate the y - the LCM of
5 and 7 is 35 - so make one coefficient 35, the other -35:)
7(2x - 5y = -4)
5(3x + 7y = 23)
14x - 35y = -28
15x + 35y = 115
29x
= 87
x = 87/29
x = 3,
(now, back-solve for y - you can use either equation: )
2(3) - 5y = -4
-6
-6
-5y = -10,
y = 2,
and the solution is the ordered pair (3,2).
(c) In this and the following problem, we're going to apply elimination to systems with
three variables. What we do here is eliminate the same variable from two of the
equations. This is best accomplished by using x with coefficient 1 (if it's available) to
eliminate the other two x's. This leaves us with a system of 2 equations, 2 variables,
which we can solve as in (a) and (b). Then we back-solve for the last variable. In the
following solutions e1 = first equation, e2 = second equation, e3 = third equation.
x + 5y + z = 7
2x + 3y - 3z = -13
(I'm going to use the first x in the system to eliminate
the other two - by multiplying e1 by -2 and adding to e2,
4x - y + 2z = -1
then *'ing e1 by -4 and adding to e3).
-2e1 + e2:
-2(x + 5y + z = 7)
2x + 3y - 3z = -13
-4e1+ e3:
-4(x + 5y + z = 7)
4x - y + 2z = -1
-2x - 10y -2z = -14)
2x + 3y - 3z = -13
-7y - 5z = -27
-4x - 20y - 4z = -28
4x - y + 2z = -1
-21y - 2z = -29
-7y - 5z = -27
-21y - 2z = -29
(I have so solve this system now by elimination - I'm going
to eliminate z - LCM(2,5) is 10 so make one coefficient on z
10, the other -10).
-2(-7y - 5z = -27)
5(-21y - 2z = -29)
14y + 10z = 54,
-105y - 10z = -145
-91y = -91
y = 1,
(back-solve for z first:)
-7(1) - 5z = -27
+7
+7
-5z = -20,
z = 4,
(now find x - I'm going to use the first equation:)
x + 5(1) +(4) = 7,
x+9=7
-9 -9
x = -2.
The solution is the ordered triple (-1,1,4).
(b) We can solve this system in the same way:
x + 2y - 3z = -7
2x – y + 4z = 11
4x + 3y – 4z = -3
-2e1 + e2:
-2(x + 2y - 3z = -7)
2x – y + 4z = 11
-2x - 4y + 6z = 14
2x - y + 4z = 11
-5y + 10z = 25
-5y + 10z = 25
-4e1+ e3:
-4(x + 2y - 3z = -7)
4x + 3y - 4z = -3
-4x - 8y + 12z = 28
4x + 3y - 4z = -3
-5y + 8z = 25
(I'm going to eliminate y)
-5y + 8z = 25
-1(-5y + 10z = 25)
-5y + 8z = 25
5y - 10z = -25
-5y + 8z = 25
2z = 0,
z = 0.
-5y + 8(0) = 25
-5y = 25,
y = -5.
(Now, back-solve for y first: )
(now, go back and get x - I'm going to use the first equation)
x + 2(-5) - 3(0) = -7
x - 10 = -7
+10 +10
x = 3.
The solution is the ordered triple (3, -5, 0)
(3)
 y  x2  2x  3
(a) 
 y  x 1
You can substitute the solution in the second equation into the first equation like this:
x - 1 = x2+2x - 3, (this is a quadratic equation - we need to move everything to one
-x +1
-x +1 side)
2
0 = x + x -2
(you can solve this by factoring or with the quadratic formula)
0 = (x + 2)(x - 1),
x + 2 = 0, x - 1 = 0,
x = -2, x = 1.
(now, we need to back-solve for the y's:)
x = -2:
y = (-2 - 1) = -3
(you can sub. x into either equation - the second is easier, though)
x = 1:
y = (1) - 1 = 0.
So the solutions are the ordered pairs (-2, -3), and (1,0).
To use the intersection method on the TI-8x, you need to enter the two equations as Y1
and Y2:
There's an x where the black square is - now graph the functions:
GRAPH
There are two points of intersection - we can find them by using the
Calc menu on the TI-83 - hit
2nd TRACE and you get the menu:
hit 5 for intersect - you will be asked what curves you want to find the
intersection of - the calculator is waiting for your response:
It wants to know if one of the curves you want to find the intersection
of is Y1 - to say yes, hit:
ENTER . Then it will ask you what the other curve is (Y2). Hit
ENTER again. Alright, now it's going to ask for a guess - the guess is important as it
tells the calculator in what area it should look for the point of intersection - let's find the
one on the left first - move the cursor back closer to it than the point of intersection on the
right:
, then ENTER , and you should get:
The first solution: (-2,-3). Now repeat this process to find the other
solution - go back to the calc menu: 2nd TRACE and hit 5 again - verify the two
curves with ENTER ENTER , and then when it asks for the guess again, move the
cursor over nearer the one on the right:
, and hit ENTER :
The second solution is the ordered pair (1,0).
x  2 y  3
(b) 
 xy  1
I'm going to solve the first equation for x (it has coefficient 1):
x - 2y = 3
2y +2y
x = 3 + 2y
(now, substitute this into the other equation for x: )
(3 + 2y)y = -1,
3y + 2y2 = -1,
(quadratic equation - move all terms to one side: )
+1 +1
2y2 + 3y + 1 = 0,
(you can factor, or use the quadratic formula: )
3  32  4(2)(1) 3  1  1


  , 1 .
2(2)
4
 2

Now we need to find the x's that go with these y's:
y = -1:
y = -1/2
x - 2(-1) = 3
x - 2(-1/2) = 3
-2
-2
-1
-1
x = 1.
x = 2.
y
The solutions are the ordered pairs (1, -1) and (2, -.5).
Note - these equations are not solved for y's, so they are actually more difficult to solve
using the graphical method, as I must solve them for y to get them into Y1 and Y2 - that
involves fractions, and is kind of messy. But feel free to give it a shot.
2
2
 2 x  y  9
(c)  2
2
 x  y  3
In this type of system, it is best to solve for x2 or y2 and substitute. I'm going to solve for
x2 in the second equation x2  y2  3
 y2  y2
x2  3  y2
Now substitute this solution into the other equation:
2(3  y 2 )  y 2  9,
6  2 y2  y2  9
-6
-6
2
3y = 3,
y2 = 1,
y = 1.
( Now, we have to find the x's that go with these y's - I'm going to use
the second equation:)
y = 1:
y = -1:
x2 - 1 = 3,
x2 - - 1 =3
+1 +1
-1 -1
x2 = 4,
x2 = 2
x =  4  2 .
x2   2 , and so we have four solutions this time:
(2,1), (2,1), ( 2, 1), and ( 2, 1).
note - this one would be really hard to do with the graphical method, so I'm going to pass.
Generally, the problems which work best with the graphical method are ones where both
equations are already solved for y.
(4) (a) note that the denominator,
x 2  8x  15  ( x  5)( x  3) in factored form, so we're going to decompose the fraction
like this here:
x
A
B
, and what we have to do is figure out what A and B are:


( x  3)( x  5) x  3 x  5
x
B 

  A
( x  5)( x  3)  2


 ( x  5)( x  3) (multiply by te LCD)
 x  8 x  15   x  3 x  5 
x  A( x  5)  B( x  3) ,
(now, we want to select x's that are zeros of
x - 5 and x - 3, namely x=5 and x=3, and plug them in this allows us to eliminate one of A or B while we solve
x = 5:
for the other: )
x  A( x  5)  B( x  3)
5  A(5  5)  B(5  3),
5  2 B,
B  5/ 2
x = 3:
x  A( x  5)  B( x  3)
5  A(3  5)  B(3  3),
5  2 A,
A  5 / 2
5
5
x
 2  2
x 2  8 x  15 x  3 x  5
5
- the denominator factors x2 - 6x = x(x - 6),
x  6x
5
A
B
(* by LCD)
 
,
x( x  6) x x  6
 5  A
B 
( x)( x  6) 
 
 ( x)( x  6),
 x( x  6)   x x  6 
5  A( x  6)  Bx ,
(pick x = 0 to knock out B, x = 6
to knock out A).
x=0:
x = 6:
5 = A(0 - 6) + B(0),
5 = A(6 - 6) + B(6),
5 = -6A,
5= 6B,
5
5
A
B .
6
6
And so,
5
5
5
 6  6 ,
x( x  6)
x
x6
(b)
2