Question

3) [3x=5y, -3x+8y=13;​

Answer 1

Відповідь:

$(x;y)=(\frac{65}{9};\frac{13}{3} )$

Пояснення:

$\left \{ {{3x=5y} \atop {-3x+8y=13}} \right.$

$x=\frac{5y}{3}$

Подставим х в -3x+8y=13

$-3*\frac{5y}{3} +8y=13\\-5y+8y=13\\3y=13\\y=\frac{13}{3}$

$x=\frac{5y}{3} \\x=\frac{5*13}{3*3} =\frac{65}{9}$

Answer 2

It looks like you have a system of linear equations:

1. 3x = 5y
2. -3x + 8y = 13

You can solve this system of equations using the method of substitution or elimination. Let's use the elimination method in this case.

First, let's multiply both sides of equation (1) by -1 to make the x coefficients cancel when we add the two equations:

1. -3x = -5y

Now, you can add equation (2) and the modified equation (1) to eliminate x:

-3x + 8y = 13 -3x = -5y

Now, add the two equations:

(-3x + 8y) + (-3x) = 13 + (-5y)

This simplifies to:

-6x + 8y = 13 - 5y

Now, combine like terms:

-6x + 8y + 5y = 13

-6x + 13y = 13

Now, isolate y by dividing both sides by 13:

(13y) / 13 = 13 / 13

y = 1

Now that we have found the value of y, you can substitute it back into equation (1) to solve for x:

3x = 5y 3x = 5(1) 3x = 5

Now, divide both sides by 3 to solve for x:

(3x) / 3 = 5 / 3

x = 5/3

So, the solution to the system of equations is x = 5/3 and y = 1.