2x-3y=6 , -3x+5y=0 find solution set by cross multiplication
5(2x-3y = 6)
3(-3x+5y = 0)
10x-15y = 30
-9x+15y = 0
now add:
x = 30
...
To solve the system of equations by cross multiplication, we need to eliminate one of the variables by multiplying each equation with an appropriate constant. Here's how it can be done:
1) Rearrange the first equation so that it is in the form "y = mx + b":
2x - 3y = 6
-3y = -2x + 6
y = (2/3)x - 2
2) Multiply the second equation by 3 to get rid of the "-3y" term:
-3x + 5y = 0
3(-3x + 5y) = 3(0)
-9x + 15y = 0
Now we have two equations in slope-intercept form and can set them equal to each other:
(2/3)x - 2 = -9x + 15y
Next, we can solve for y in terms of x:
15y = (2/3)x - 2 + 9x
15y = (2/3)x + 9x - 2
15y = (29/3)x - 2
Finally, divide both sides of the equation by 15 to isolate y:
y = (29/45)x - 2/15
So, the solution set for the given system of equations is:
{(x, y) | y = (29/45)x - 2/15}