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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Hadley 1 month 2021-08-12T09:55:44+00:00 1 Answer 0 views 0

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    2021-08-12T09:57:24+00:00

    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

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