Question

6x-4у - 12=03x + y -15=0​

Answer 1

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Answer 2

It looks like you've provided a system of two linear equations with two variables, x and y. The system of equations is as follows:

1. $6x - 4y - 12 = 0$
2. $3x + y - 15 = 0$

To solve for the values of x and y that satisfy both equations simultaneously, we can use different methods, such as substitution or elimination. I'll use the substitution method here:

Let's solve equation (2) for y: $3x + y = 15$ $y = 15 - 3x$

Now, substitute this value of y into equation (1): $6x - 4(15 - 3x) - 12 = 0$ $6x - 60 + 12x - 12 = 0$ $18x - 72 = 0$ $18x = 72$ $x = \frac{72}{18}$ $x = 4$

Now that we have the value of x, we can substitute it back into the equation for y: $y = 15 - 3x$ $y = 15 - 3(4)$ $y = 15 - 12$ $y = 3$

So, the solution to the system of equations is $x = 4$ and $y = 3$. Both equations are satisfied when these values are plugged in.