## 3) Solve the equations by substitution method 3x – y = 23; 4x + 3y = 48. 4) Given 4a + 3b = 65 and a + 2b 35 solve by elimination metho

Question

3) Solve the equations by substitution method 3x – y = 23; 4x + 3y = 48.
4) Given 4a + 3b = 65 and a + 2b 35 solve by elimination method.
5) Solve 2x = -7y + 5; -3x = -8y – 11 by the method of cross multiplication.​

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11 months 2021-08-12T09:55:44+00:00 1 Answer 0 views 0

1. 4a+3b=65…(i)

4(a+2b=35)

=> 4a+8b=140…(ii)

Subtract (ii) from (i),

-5b=-75

b=15, a=5 for question 4

Given 2 Equations That:-

3x – y = 23 —— (i)

4x + 3y = 48 —— (ii)

consider equation (i)

3x – y = 23 x 3

= 9x – 3y = 69 ———- (iii)

consider equation (ii) and (iii)

9x – 3y = 69

4x + 3y = 48

13x = 117

x = 117/13

Therefore, x = 9

Substitute x = 9 in (ii)

4x + 3y = 48

4(9) + 3y = 48

36 + 3y = 48

3y = 48 – 36

3y = 12

y = 12/3

Therefore, y = 4

therefore, x = 9 , y = 4

X=9 , Y =4

Step-by-step explanation:

elimination method

given 2 equations that

3x-y = 23……….(1)

4x+3y= 48………(2)

consider equation ……..(1)

3x-y = 23×3

or,9x-3y =69 …………(3)

consider equation (2) and (3)

9x-3y = 69

4x+3y = 48

____________

13x = 117

or, x = 117/13

therefore, x=9

substitute, x=9 in (2)

4x+3y = 48

4(9)+3y= 48

36+3y =48

or, 3y = 48-36

or, 3y= 12

or, y= 4

therefore , x= 9 , y = 4 for question 3

i don’t know how to solve the 5th question i ami really sorry. hope you understand