Question

Find the value of a and b for which each of the following system of equation has many solution (-1)x + 3y 2. 6x + (1- 26)y 6 (h) (2a - 1)x+ 3y 5.3x + 9(1 - 1)y 2​

Answer 1

Consider the system x − 3y = 6

−2x + 5y = −5

of linear equations. In this case a solution to the

system is an ordered pair (x, y) that makes both equations true. A large part of linear algebra

concerns itself with methods of solving such systems, and ways of interpreting solutions or lack of

solutions.

In the past you should have learned two methods for solving such systems, the addition

method and the substitution method. The method we want to focus on is the addition method.

In this case we could multiply the first equation by two and add the resulting equation to the sec-

ond. (The first equation itself is left alone.) The result is x − 3y = 6

−y = 7 ; from this we can see

that y = −7. This value is then substituted into the first equation to get x = −15.

Sometimes we have to do something a little more complicated:

⋄ Example 1.2(a): Solve the system 2x − 4y = 18

3x + 5y = 5 using the addition method.

Here we can eliminate x by multiplying the first equation by 3 and the second by −2, then

adding:

2x − 4y = 18

3x + 5y = 5 =⇒

6x − 12y = 54

−6x − 10y = −10

−22y = 44

y = −2

Now we can substitute this value of y back into either equation to find x:

2x − 4(−2) = 18

2x + 8 = 18

2x = 10

x = 5

The solution to the system is then x = 5, y = −2, which we usually write as the ordered pair

Answer 2

Consider the system x - 3y = 6

-2x + 5y = -5

of linear equations. In this case a solution to the

system is an ordered pair (x, y) that makes both equations true. A large part of linear algebra

concerns itself with methods of solving such systems, and ways of interpreting solutions or lack of

solutions.

In the past you should have learned two methods for solving such systems, the addition

method and the substitution method. The method we want to focus on is the addition method.

In this case we could multiply the first equation by two and add the resulting equation to the sec

ond. (The first equation itself is left alone.) The result is x-3y = 6

-y = 7; from this we can see

that y=-7. This value is then substituted into the first equation to get x = -15.

Sometimes we have to do something a little more complicated:

Example 1.2(a): Solve the system 2x - 4y = 18

3x + 5y = 5 using the addition method.

Here we can eliminate x by multiplying the first equation by 3 and the second by-2, then

2x - 4y = 18

3x + 5y = 5 =>

6x - 12y = 54

-6x - 10y = -10

-22y = 44 =

y=-2

Now we can substitute this value of y back into either equation to find x:

2x 4(-2) = 18 =

2x + 8 = 18

2x = 10

X = 5

The solution to the system is then x =5 v=-2.