You can't. All you can tell is the relationship between 'x' and 'y'.
x = 40 - 0.75y
y = 1/3 of (160 - 4x)
There are an infinite number of pairs of numbers that 'x' and 'y' can be.
In order to find out a single pair of numbers that 'x' and 'y' must be, you need another equation.
You need the same number of equations as the number of 'unknowns' you're trying to find.
If: 30y = 10+y Then: y = 10/29
15z +5x-30y = 5(x - 6y + 3z)
Oh, dude, it's like super easy. You just like, factor out the common factors, which in this case are 5 and the variables x and y. So, the factored form would be 5xy(6). Boom, math problem solved!
3x+5y=48 5y=48-3x-3x+5y=12 -3x+(48-3x)=12-6x=-36x=65y=48-3(6)5y=30y=6(6,6)
9x-6y-(12*2y)=0 9x-6y-24y=0 9x-30y=o 9x=30y 3x=10y y=0.3x x=10/3y
9y^-30y^ + 24y^3 = 2y^3(3y - 4)(y - 2)
-30y
12y3 - 30y2 + 12yFactor out a y:y(12y2 - 30y + 12)Factor out a 6:6y(2y2 -5y + 2)Factor 2y2 - 5y + 26y(2y - 1)(y - 2)
3x + 9 y = 120 5x + 5y = 90 multiply first equation by 5 ( both sides) multiply second equation by 3 ( both sides) 15 x + 45 y = 600 15 x + 15 y = 270 subtract 30y = 330 divide by 30 y = 11 substiute y into first equation 3 x + 9 y = 120 3x + 99 = 120 3x = 21 x = 7
Rearrange the second equation as x = 10+y and then substitute it into the first equation which will create a quadratic equation in the form of: 2y2+30y+100 = 0 and when solved y = -10 or y = -5 Therefore the solutions are: x = 0, y = -10 and x = 5, y = -5
The GCF is 6.
If: x^2+y^2+4x+6y -40 = 0 and x -y = 10 Then by rearranging: x = 10+y and 2y^2+30y+100 = 0 Solving the above quadratic equation: y = -10 and y = -5 Points of intersection by substitution are: (0, -10) and (5, -5)