4y-3 = 10y+45

Let's change y to x and graph it

4x-3 = 10x+45

Graph each side and see where they intersect

Left side has slope = 4 with y-intercept of -3

right side has slope of 10 and y-intercept of 45

Just a rough sketch shows they intersect where x is negative.

the right side is steeper and crossing the y-axis higher. It can never intersect the smaller sloped line with the smaller y intercept

the solution is a negative number

Graph it more carefully and you can get the exact solution of -8 or the point (-8,-35). The two lines intersect in Quadrant III. since we switched x and y, y=-8 is the solution to the original problem.

You can do the algebra

4y-3 = 10y+45

subtract 45 from both sides, subtract 4y from both sides to get

-48=6y

divide both sides by 6

-48/6 = -8 = y

y=-8

check the answer

4(-8)-3 = 10(-8)+45

-32-3 = -80+45

-35 = -35