Math, asked by rajnandinis810, 3 months ago

8x+12y=3and 2x+6y=-1 solve by substution method​

Answers

Answered by Legend42
36

Answer:

functions of the same independent variable. Consider the prior example where x, the number of items produced and sold, was the independent variable in three functions, the cost function, the revenue function, and the profit function.

In general there may be:

n equations

v variables

Solving systems of equations

There are four methods for solving systems of linear equations:

a. graphical solution

b. algebraic solution

c. elimination method

d. substitution method

Graphical solution

Example 1

given are the two following linear equations:

f(x) = y = 1 + .5x

f(x) = y = 11 - 2x

Graph the first equation by finding two data points. By setting first x and then y equal to zero it is possible to find the y intercept on the vertical axis and the x intercept on the horizontal axis.

If x = 0, then f(0) = 1 + .5(0) = 1

If y = 0, then f(x) = 0 = 1 + .5x

-.5x = 1

x = -2

The resulting data points are (0,1) and (-2,0)

Graph the second equation by finding two data points. By setting first x and then y equal to zero it is possible to find the y intercept on the vertical axis and the x intercept on the horizontal axis.

If x = 0, then f(0) = 11 - 2(0) = 11

If y = 0, then f(x) = 0 = 11 - 2x

2x = 11

x = 5.5

The resulting data points are (0,11) and (5.5,0)


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Answered by StrangeStinkle
9

Answer:

(i) x + y =5 and 2x –3y = 4

By elimination method

x + y =5 ... (i)

2x –3y = 4 ... (ii)

Multiplying equation (i) by (ii), we get

2x + 2y = 10 ... (iii)

2x –3y = 4 ... (ii)

Subtracting equation (ii) from equation (iii), we get

5y = 6

y = 6/5

Putting the value in equation (i), we get

x = 5 - (6/5) = 19/5

Hence, x = 19/5 and y = 6/5

By substitution methodx + y = 5 ... (i)

Subtracting y both side, we get

x = 5 - y ... (iv)

Putting the value of x in equation (ii) we get

2(5 – y) – 3y = 4

-5y = - 6

y = -6/-5 = 6/5

Putting the value of y in equation (iv) we get

x = 5 – 6/5

x = 19/5

Hence, x = 19/5 and y = 6/5 again

(ii) 3x + 4y = 10 and 2x – 2y = 2

By elimination method

3x + 4y = 10 .... (i)

2x – 2y = 2 ... (ii)

Multiplying equation (ii) by 2, we get

4x – 4y = 4 ... (iii)

3x + 4y = 10 ... (i)

Adding equation (i) and (iii), we get

7x + 0 = 14

Dividing both side by 7, we get

x = 14/7 = 2

Putting in equation (i), we get

3x + 4y = 10

3(2) + 4y = 10

6 + 4y = 10

4y = 10 – 6

4y = 4

y = 4/4 = 1

Hence, answer is x = 2, y = 1

By substitution method

3x + 4y = 10 ... (i)

Subtract 3x both side, we get

4y = 10 – 3x

Divide by 4 we get

y = (10 - 3x )/4

Putting this value in equation (ii), we get

2x – 2y = 2 ... (i)

2x – 2(10 - 3x )/4) = 2

Multiply by 4 we get

8x - 2(10 – 3x) = 8

8x - 20 + 6x = 8

14x = 28

x = 28/14 = 2

y = (10 - 3x)/4

y = 4/4 = 1

Hence, answer is x = 2, y = 1 again.

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

By elimination method

3x – 5y – 4 = 0

3x – 5y = 4 ...(i)

9x = 2y + 7

9x – 2y = 7 ... (ii)

Multiplying equation (i) by 3, we get

9 x – 15 y = 11 ... (iii)

9x – 2y = 7 ... (ii)

Subtracting equation (ii) from equation (iii), we get

-13y = 5

y = -5/13

Putting value in equation (i), we get

3x – 5y = 4 ... (i)

3x - 5(-5/13) = 4

Multiplying by 13 we get

39x + 25 = 52

39x = 27

x =27/39 = 9/13

Hence our answer is x = 9/13 and y = - 5/13

By substitution method

3x – 5y = 4 ... (i)

Adding 5y both side we get

3x = 4 + 5y

Dividing by 3 we get

x = (4 + 5y )/3 ... (iv)

Putting this value in equation (ii) we get

9x – 2y = 7 ... (ii)

9 ((4 + 5y )/3) – 2y = 7

Solve it we get

3(4 + 5y ) – 2y = 7

12 + 15y – 2y = 7

13y = - 5

y = -5/13

x = 4 + 5 ( -5/13)/ 3

= 4 - 25/13 / 3

= 4 × 13 - 25/13 / 3

= 27/13×3

= 27/39

= 9/13

Hence we get x = 9/13 and y = - 5/13 again.

(iv) x/2 + 2y/3 = - 1 and x – y/3 = 3

By elimination method

x/2 + 2y/3 = -1 ... (i)

x – y/3 = 3 ... (ii)

Multiplying equation (i) by 2, we get

x + 4y/3 = - 2 ... (iii)

x – y/3 = 3 ... (ii)

Subtracting equation (ii) from equation (iii), we get

5y/3 = -5

Dividing by 5 and multiplying by 3, we get

y = -15/5

y = - 3

Putting this value in equation (ii), we get

x – y/3 = 3 ... (ii)

x – (-3)/3 = 3

x + 1 = 3

x = 2

Hence our answer is x = 2 and y = −3.

By substitution method

x – y/3 = 3 ... (ii)

Add y/3 both side, we get

x = 3 + y/3 ... (iv)

Putting this value in equation (i) we get

x/2 + 2y/3 = - 1 ... (i)

(3+ y/3)/2 + 2y/3 = -1

3/2 + y/6 + 2y/3 = - 1

Multiplying by 6, we get

9 + y + 4y = - 6

5y = -15

y = - 3

Hence our answer is x = 2 and y = −3

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