# Need help with this kind of math problems?

How to solve this kind of math problems
65. Y=2x
Y=2x+3

66. Y= 3\5 x -10
Y= 5/3x-10

67. 10x+5y=30
Y= -2x+6

68. 3x+8y=20
3x+6y=14

69. 2x-3y=19
5x-y=28

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68)  3x + 8y = 20------(1)
3x + 6y = 14------(2)

Subtract (1) and (2) eqations to get y value and group like terms

3x - 3x + 8y - 6y = 20 - 14

⇒ 2y = 6

⇒ y = 3

Now substitute y = 3 in eqation (1) to get x value

3x + 8 (3) = 20

⇒ 3x + 24 = 20

⇒ 3x + 24 - 24 = 20 -24

⇒ 3x = - 4

⇒ x = -4/3

The values of x is -4/3 and y is 3

67 ) Given two equations

10x + 5y = 30    y = -2x + 6

Let 10x + 5y = 30 ----------------->(1) and

2x + y = 6 ----------------->(2)

Now multiply 5(2)

Then,             10x + 5y = 30

10x + 2y = 12

( - )_____________________

3y = 18

y = 6

Substitute y = 6 in equation (1)

10x + 5(6) = 30

10x + 30 = 30

10x = 0

x = 0

The values of  x = 0 and y = 6

There are infinitely many solutions.

66. Y= 3\5 x -10
Y= 5/3x-10

Given equations y = 3 / 5 x -10; y = 5/3 x - 10

y = 3 / 5 x -10-------(1)

y = 5/3 x - 10-------(2)

from  equation (1) and equation (2)

3/5x-10 = 5/3x -10

Subtract 5/3 x from each side

3/5 x -10 - 5/3x = 5/3 x -10 -5/3 x

3/5 x -5/3 x -10 = -10

3/5 x-5/3 x-10+10 = -10+10

x ( 3/5 -5/3) = 0

Divide each side by ( 3/5-5/3)

x( 3/5-5/3) / ( 3/5-5/3) = 0 / ( 3/5-5/3)

x=0

x=0 substitute in equation y = 3 / 5 x -10

y = 3/5 (0) -10

y = -10

The values of x=0 ;y= -10

69. 2x-3y=19
5x-y=28

Given two equations

2x - 3y = 19;5x- y  = 28

Let2x - 3y = 19 ----------------->(1) and

;5x- y  = 28 ----------------->(2)

Now multiply equation (2) by 3

Then,             2x-3y = 19

(-) 15x-3y = 84

_____________________

-13x = -65

Divide each side by negtive 13

-13x / -13 =  - 65/ -13

x=5

Substitute x = 5 in equation (1)2x - 3y = 19

2(5) - 3y = 19

10 - 3y = 19

Subtract10 from each side

10 -3y  -10 = 19-10

-3y = 9

Divide each side by negtive 3

-3y/-3 = 9 / -3

y = -3

The values of x=5 ;y= -3

65.   Y = 2x
Y = 2x + 3

Here the coefficients of x and y are equal

So, we can`t solve these equqtions for x and y

67) 10x + 5y = 30            -> (1)

y = -2x  + 6                    -> (2)

Substitute y = -2x  + 6 in equation (1).

10x + 5y = 30

10x + 5(-2x  + 6) = 30

Apply Distributive Property: a(b + c) = ab + ac.

10x + 5(-2x) + 5 (6) = 30

10x - 10x + 30 = 30

30 = 30

This is a true statement.

Therefore there are infinitely many solutions.

Let x  = a where a Є R.

Therefore (x , y )(a, -2a + 6) where a Є R.

65) y = 2x                    -> (1)

y = 2x  + 3                   -> (2)

Substitute y = 2x  in equation (2).

2x = 2x  + 3

Subtract 2x  from each side.

2x - 2x = 2x - 2x  + 3

0 = 3

The statement is false. So, there is no solution.