# LINEAR AND QUADRATIC EQUATION SYSTEMS

## LINEAR AND QUADRATIC EQUATION SYSTEMS

### SYSTEM OF LINEAR EQUATIONS IN TWO VARIABLES

The general form of linear equations in two variables is :

a1x + b1y = c1              , a1 ,b1 not both 0

a2x + b2y = c2              , a2 ,b2 not both 0

Solution set of linear equations in two variables can be found by :

1.      Graphs

2.      Elimination

3.      Substitution

4.      Combination of elimination and substitution

1.      Solution by Graphs

To solve a system of linear equations in variable x and y by graphs, we must draw the two equations on the same coordinate system. Then find the point of intersections them. This point is called the solution of linear equations in two variables.
Examples :
1. By graph, find the solution set of        2x + y = 4

x – y = -1

Solution :

( Please, do it in millimeter paper)

Step 1. Draw the two equations on the same coordinate system

If   , then system of linear equations in two variables have only one solution and the graphs intersect in one and only one point.

2. By the graph, find the solution set of    x + 2y = 4

Step 2 . The system have no solution, because the graphs have no intersection point

Conclude :

If   , then system of linear equations in two variables have no solution

And the graphs are distinct parallel lines.

3. By the graph, find the solution set of         x – y = 2

2x – 2y = 4

Solution :

Step 1.         x – y = 2                                  2x – 2y = 4

x                                              x

y                                              y

The graph:

Step 2. The system have infinitely many solutions, because the graphs have many

Intersection point

Conclude :

If  , then system of linear equations in two variables have infinitely

Many solutions and the graphs are the same line.

2. Solution by Elimination

There are two steps :

a.       eliminate one variable

b.      eliminate other variable

Examples :

1. By elimination, find the solution set of              2x – y = 4

3x + 2y = 13

Solution :

2x – y = 4 …………..(1)

3x + 2y = 13………...(2)

Elimination of y:

4x – 2y = 8………….(3) ( multiply (1) by 2)

3x + 2y = 13

+

7x = 21                (addition of  (3) and (2) so that eliminate the variable y)

x = 3

Elimination of x :

6x – 3y = 12…………(4) (multiply (1) by 3)

6x + 4y = 26…………(5) (multiply (2) by 2)

-7y = -14                     (subtraction (4) and (5) so that eliminate the

y =  2                             variable x)

# The solution set is

2. By elimination, find the solution set of              2x + y = 4

x – y  = -1

Solution :

2.      Solution by Substitution

There are two steps :

a.       Take one equation and express one variable with other variable

b.      Substitute to other equation

Examples :

1. By substitution, find the solution set of           x + y = 4

4x + 3y = 13

Solution :

x + y = 4……..(1)       y = 4 – x …….(3) ( express y with x)

4x + 3y = 13 ……(2)

Substitute (3) to (2)

4x + 3y = 13

4x + 3(4- x)= 13

4x + 12 – 3x = 13

x = 1

Substitute x = 1 to (3)

y = 4 – x

= 4 – 1

= 3

# The solution set is

2. By substitution, find the solution set is    2x + y = 8

3x + 2y = 13

Solution :

4. Solution by Combination of Elimination and Substitution

Examples :

1) By combination of elimination and substitution, find the solution set of

4x + 3y = 10

2x + y   =   4

Solution :             4x + 3y = 10    x 1      4x + 3y = 10

2x + y   =   4     x 3      6x + 3y = 12

-

- 2x  = - 2

x  =  1

Substitute     x = 1 to equation which easy, example :  2x + y = 4

2.1 + y = 4

y = 2

☻ The solution set  is

2) By combination of elimination and substitution, find the solution set of

Solution :

EXERCISES :

1.  Find the solution set of the following linear equation systems by graphs

a.     3x –y = 2              b.     2x + 8y = 6                      c.      3x – y = 3

x + y = 2                     4x + 6y = 12                             6x – 2y = 12

2.  Find the solution set of the following linear equation systems by elimination

a.      2x + 11y = -1                  b.      5x = 2y + 4                     c.      –x + 2y + 1 = 0

3x – 7y  = 22                           x – 4y = -10                            2x + 3y + 3 = 0

3.  Find the solution set of the following linear equation systems by substitution

a.               b.                                 c.

4.  Find the solution set of the following linear equation systems by combination of elimina-

tion and substitution

a.                                   d.            g.

b.               e.           h.           c.                     f.                   i

5.  Find point intersection of the lines  y = 2x – 2  and x – y +1 = 0 and sketch it

6.  Given three resistors, they are resistor   A = R1 ohms, resistor B = R2 ohms and resistor

C = R2 ohms. When resistors A and B are serried, the shunt resistor is 10 ohms. When

Resistors A, B, and C are serried the shunt resistor is 13 ohms. Find shunt resistor when

Resistors A, B, and C are parallelized

7. Mr. Agus works for 6 days which  4 days are overtime to get  74,000 rupiah. Mr. Bardi

works for 5 days which 2 days are overtime to get 55,000 rupiah. Mr. Agus, Mr. Bardi,

and Mr. Dodo work under the same payment system. If Mr. Dodo works for 5 days over-

time, then the payment that he shall receive is ….

8.  In the year 2002 the age of a girl is equal to a quarter of her mother’s age. In the year

2006 the age of the girl is one third her mother’s age. Find the year when the girl born

9.  If the numerator of fraction is added 2 and the denominator is added 5 then the value of

fraction is a half. If the numerator minus one and the denominator is added one then the

value of fraction is a third. Find the fraction.

10. The price of eggs and meat in two shops are shown in the following table :

Eggs (kg)

Meat (kg)

Price ( Thousand rupiahs )

Shop A

100

200

6900

Shop B

60

70

2640

Find the price of one kg egg and one kg meat in two shops

B.  SYSTEM OF LINEAR EQUATIONS IN THREE VARIABLES

The general form is :

a1x + b1y + c1z = d1      ;  a1, b1, c1  not both 0

a2x + b2y + c2z = d2      ;  a2, b2, c2  not both 0

a3x + b3y + c3z = d3      ;  a3, b3, c3  not both 0

Solution set of linear equations in three variables can be found by :

a.       Substitution

b.      Combination of elimination and substitution

1. Solution by substitution

Example :

Find the solution set of

Solution :

Equation (1) is made become  y = 7 – 2x – 3z …………(4)

Substitute equation (4) to equation (2) so  :

x + 2 ( 7 – 2x – 3z ) + z = 1

x + 14 – 4x – 6z + z = 1

- 3x – 5z = - 13

3x + 5z = 13 …………..(5)

Substitute equation (4) to equation (3) so :

……………………………………………..

……………………………………………..

……………………………………………..

z = 3 – x ……….(6)

Substitute equation (6) to equation (5) so  :

………………………………………………

………………………………………………

………………………………………………

x = ….

Substitute x =…  to equation (6) so :   z = ….

Substitute x = … and   z = ….  To equation (4) so  y = ….

☻Solution set is

2. Solution set by  combination of substitution and elimination

Example :

Find solution set of

Solution :

Elimination z of equation (1) and (2)

2x + y –z = 5

x + 2y +2z = 13

………………..

…………………………………………(4)

Elimination z of equation (1) and (3)

2x +y –z = 5                 x 2       …………………………

x – 3y + 2z = -4           x1        …………………………

+

…………………………...............(5)

Elimination y of equation (4) and (5)

…………………..

…………………..

-

……………………

x = …..

Substitute x = …. To equation (4)

……………………..

……………………..

y = ….

Substitute x = … and y = … to equation (1)

……………………..

……………………..

z = ….

☻The solution set is

EXERCISES

1.  Find the solution set of the following linear equation systems by substitute

a.                  b.                          c.

2.  Find the solution set of the following linear equation systems by combination of substitution

and elimination

a.                   b.                        c.

d.                    e.

3.  The price of  two bananas, two guavas, and one mango is 1,400 rupiahs. The price of one

banana, one guava, and two mangoes is 1,300 rupiahs. The price of one banana, three

guavas, and one mango is 1,500 rupiahs. Find the price of  one banana, one guava and one

mango.

4.  The graph of quadratic function y = ax2 + bx + c passes through the points (-1,0),(1,6) and

(2,12). Find a, b, c, then give the equation of the graph.

5.  The circle  x2 + y2 + Ax + By +C = 0, passes through the points (3,-1),(5,3) and (6,2). Find

A, B, C and then give the equation.

6.  The quadratic form ax2 + bx + c has value -1 if x = 1, its value 4 if x = 2, and has value 17

if x = 3. Find value a, b, and c

C. SIMULTANEOUS EQUATIONS, ONE LINEAR – ONE QUADRATIC.

1.      Solution set by substitute

Example :

a)  Find the solution set of

Solution :

Substitute equation (1) to (2)

x2 +1 = x + 3

…………………….

…………………….

x = ….     or   x = ….

For  x = ….,   then  y = …..        and the point is (….,….)

For x = ….,    then y = …..         and the point is (….,….)

☻Solution set is

NOTE :

The solution set is the point intercept of (1) and (2).

By the graph can be expressed as follows :

Y

y = x2 + 1

y = x + 3

(2,5)

(-1,2)

X

b)  Find the solution set of

Solution :

2x2 – 3x + 1 = x2 + x + 6

x2 – 4x – 5 = 0

(x + 1)(x – 5) = 0

x = -1  or  x = 5

For  x = -1 then  y = 6

For  x = 5   then y = 36

☻Solution set is

2. Solution set by factoring of the quadratic form

Example :

Find the solution set of

Solution :

Factorize the  equation (2)

x2 + 2xy + y2 -25 = 0

(x + y + 5)(x + y – 5) = 0

x + y + 5 = 0    or   x + y – 5 = 0

For  x + y + 5 = 0 then  y = - x – 5

Substitute y = - x – 5  to  (1) :

(- x – 5) – x = 3

-2x = 8

x = -4   then   y = -1,  so the point is (-4,-1)

For  x + y – 5 = 0  then  y = 5 – x

Substitute  y = 5 – x to (1) :

(5 – x) – x = 3

-2x = -2

x = 1   then  y = 4,  so the point is (1,4)

☻Solution set is

EXERCISES

1. Find the solution set of the following equation system :

a.              d.

b.                           e.

c.                                     f.

2. Find the border value of m so that the line y = x – 10 and curve y = x2 – mx + 6 have two

distinct intercept point

3. Find value of m so that  the line y = mx – 9  touches the quadratic function y = x2

4. Given the equation system

Find the border value of m so that  the equation system has two point in solution