LINEAR EQUATIONS WITH UNKNOWN COEFFICIENTS WORKSHEET

Problem 1 :

For what value of c will system of equations below have no solution ?

cx – 2y  =  6

3x + 4y  =  4

Solution

Problem 2 :

For what value of b will system of equations below have infinitely many solutions ?

-2x + y  =  4

5x - by  =  -10

Solution

Problem 3 :

ax - y  =  0

x - by  =  1

In the system of equations above, a and b are constants and x and y are variables. If the system of equation above has no solution, what is the value of a . b ?

Solution

Problem 4 :

2x - ky  =  14

5x - 2y  =  5

In the system of equations above, k is a constants and x and y are variables. For what value of k will the system of equations have no solution ?

Solution

Example 5 :

2x – 1/2y  =  15

ax – 1/3y  =  10

In the system of equations above, a is a constants and x and y are variables. For what value of a will the system of equations infinitely many solutions ?

Solution

Problem 6 :

ax + 4y  =  14

5x + 7y  =  8

In the system of equations above, a is a constants and x and y are variables. If the system no solution, what is the value of a ?

Solution

Problem 7 :

ax + 1/2y  =  16

4x + 3y  =  8

In the system of equations above, a is a constants and x and y are variables. If the system no solution, what is the value of a ?

Solution

Problem 8 :

3x + ky  =  8

x + 4y  =  -1

If(x, y) is a solution to the system of equations above and k is a constant, what is y in terms of k ?

Solution

Problem 9 :

5x + 16y  =  36

cx + dy  =  9

The system of equations above, where c and d are constants, has infinitely many solutions. what is the value of cd ?

Solution

Problem 10 :

0.3x – 0.7y  =  1

kx – 2.8y  =  3

In the system of equations above, k is a constant. If the system no solution, what is the value of k ?

Solution

(1)  c  =  -3/2

(2)  b  =  5/2

(3)  a.b  =  1

(4)  k  =  4/5

(5)  a  =  4/3

(6)  a  =  20/7

(7)  a  =  2/3

(8)  5/(k – 12)

(9)  d  =  4/5 and c  =  1/4

(10)  k  =  1.2

Problem 1 :

Find constants a, b and c given that

2x² + 4x + 5 = ax² + [2b - 6]x + c for all x

Solution

Problem 2 :

2x³ - x² + 6 = (x - 1)² (2x + a) + bx + c for all x.

Solution

Find a and b if :

Problem 3 :

z⁴ + 4 = (z² + az + 2) (z² + bz + 2) for all z.

Solution

Problem 4 :

2z⁴ + 5z³ + 4z² + 7z + 6 = (z² + az + 2) (2z² + bz + 3) for all z.

Solution

Problem 5 :

Show that z⁴ + 64 can be factorised into two real quadratic factors of the form z² + az + 8 and z² + bz + 8, but cannot be factorised into two real quadratic factors of the form z² + az + 16 and z² + bz + 4.

Solution

Problem 6 :

Find real numbers a and b such that

x⁴ - 4x² - 8x - 4 = (x² + ax + 2) (x² + bx - 2)

and hence solve the equation. x⁴ + 8x = 4x² + 4.

Solution

Problem 1 :

2z – 3 is a factor of 2z³ – z² + az – 3. Find a and all zeros of the cubic.

Solution

Problem 2 :

3z + 2 is a factor of 3z³ – z² + [a + 1]z + a. Find a and all the zeros of the cubic.

Solution

Problem 3 :

Both 2x + 1 and x – 2 are factors of P(x) = 2x⁴ + ax³ + bx² – 12x – 8. Find a and b and all zeros of P(x).

Solution

Problem 4 :

x + 3 and 2x – 1 are factors of 2x⁴ + ax³ + bx² + ax + 3. Find a and b and hence determine all zeros of the quartic.

Solution

Problem 5 :

a. x³ + 3x²  - 9x + c has two identical linear factors. Prove that c is either 5 or -27 and factories the cubic into linear factors in each case.

Solution

Problem 6 :

3x³ + 4x² – x + m has two identical linear factors. Find m and find the zeros of the polynomial in all possible cases.

Solution

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