Question

2x+3y=0 and 3x+4y=5​

Answer 1

Answer:

$2x + 3y = 0 \\ = > 2x = - 3y \\ = > x = - \frac{3y}{2} \: \: \: \: ........(1) \\ \\ 3x + 4y = 5 \\ = > 3 \times ( - \frac{3y}{2} ) + 4y = 5 \\ = > - \frac{9y}{2} + 4y = 5 \\ = > \frac{ - 9y + 8y}{2} = 5 \\ = > \frac{ - y}{2} = 5 \\ = > - y = 5 \times 2 \\ = > y = - 10 \: \: ans \\ \\ putting \: the \: values \: of \: y \: in \: equation \: (1) \\ x = \frac{ - 3y}{2} \\ = > x = \frac{ - 3 \times ( - 10)}{2} \\ = > x = \frac{30}{2} \\ = > x = 15$

Answer 2

Answer:

Y=-10 X=15

Step-by-step explanation:

Using Elimination Method

2x+3y=0 X3        (Coefficient of x from 2nd equation)
3x+4y=5 X2        (Coefficient of x from 1st equation)

6x+9y=0
6x+4y=5
Cut 6x because its equal
Now change in sign (+ becomes - and - becomes +) of the second equation
+9y=0
-8y=-10
After solving we get
1y=-10
y=-10
now substitute -10 into 1st equation ( Y variable )
2x+3(-10) = 0 ( Multiplication )

2x+30y = 0
2x=-30 ( 30y will shift to constant since the constant is 0 there is no need of substracting )

30÷2 x= 15
Hope it helps you