To solve this system of equations, we can use the method of substitution or elimination.
Let's use the elimination method.
First, we can simplify the first equation by dividing by 3:5x - 3y = 8(5/3)x - y = 8/3
Now, let's multiply the second equation by 1/3 to match the coefficient of y in the first equation:15x -9y = 8(15/3)x - 3y = 8/3
Now, we can subtract the first equation from the second equation to eliminate y:(15/3)x - 3y - (5/3)x + y = 8/3 - 8/3(10/3)x - 2y = 0
Now we have a new equation:(10/3)x - 2y = 0
To solve for x, we can isolate x in terms of y:(10/3)x = 2yx = (3/10) * 2yx = 6y/10x = 3y/5
Now we can substitute this value of x back into the first equation to solve for y:5(3y/5) - 3y = 83y - 3y = 80 = 8
Since this equation is not true, it means that the system of equations is inconsistent and there is no solution.
To solve this system of equations, we can use the method of substitution or elimination.
Let's use the elimination method.
First, we can simplify the first equation by dividing by 3:
5x - 3y = 8
(5/3)x - y = 8/3
Now, let's multiply the second equation by 1/3 to match the coefficient of y in the first equation:
15x -9y = 8
(15/3)x - 3y = 8/3
Now, we can subtract the first equation from the second equation to eliminate y:
(15/3)x - 3y - (5/3)x + y = 8/3 - 8/3
(10/3)x - 2y = 0
Now we have a new equation:
(10/3)x - 2y = 0
To solve for x, we can isolate x in terms of y:
(10/3)x = 2y
x = (3/10) * 2y
x = 6y/10
x = 3y/5
Now we can substitute this value of x back into the first equation to solve for y:
5(3y/5) - 3y = 8
3y - 3y = 8
0 = 8
Since this equation is not true, it means that the system of equations is inconsistent and there is no solution.