Answer:-₍1,3₎
Explanation:
To solve this system of equations, we can use the method of elimination to eliminate one of the variables. We can multiply the first equation by 5 and the second equation by -3, then add the two equations to eliminate $y$:
$(5)(2x + 3y = 4) \Rightarrow 10x + 15y = 20$
$(-3)(3x + 5y = 7) \Rightarrow -9x - 15y = -21$
Adding the equations, we get:
$10x + 15y - 9x - 15y = 20 - 21$
Simplifying, we get:
$x = -1$
Now we can substitute $x=-1$ into one of the original equations to solve for $y$. Using the first equation, we have:
$2(-1) + 3y = 4$
Solving for $y$, we get:
$y = 2$
Therefore, the solution to the system of equations is $x=-1$ and $y=2$. We can check this solution by substituting $x=-1$ and $y=2$ into the other two equations:
$3(-1) + 5(2) = 7$
$10(-1) + 15(2) = 20$
Both equations are true, so our solution is correct.