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.
Answers
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