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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.
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What is y if 30y equals 10 plus y?
If: 30y = 10+y Then: y = 10/29
How do you factor 15z plus 5x-30y?
15z +5x-30y = 5(x - 6y + 3z)
3x plus 5y equals 48 -3x plus 5y equals 12Solve the system of equations Enter your answer as an ordered pair?
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 x 2y0 How do you solve using elimination using multiplication?
9x-6y-(12*2y)=0 9x-6y-24y=0 9x-30y=o 9x=30y 3x=10y y=0.3x x=10/3y
How do you factor 9y5-30y4 plus 24y3?
9y^-30y^ + 24y^3 = 2y^3(3y - 4)(y - 2)
What is 5y-35?
-30y
How do you factor 12y to the third power minus 30y squared plus 12y?
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)
How do you solve the simultaneous equation 3x plus 9y equals 120 5x plus 5y equals 90?
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
What are the solutions to the simultaneous equations of x square plus y square plus 4x plus 6y minus 40 equals 0 and x minus y equals 10?
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
What is the GCF of 12x and 30y?
The GCF is 6.
What are the points of intersection of x2 plus y2 plus 4x plus 6y -40 equals 0 with x -y equals 10 showing key stages of work?
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)
What is the solution of the system of equations 3x plus 6y and ndash2 15x plus 30y and ndash10?
Without any equality signs the given expressions can't be considered to be simultaneous equations and so therefore no solutions are possible.
