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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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∙ 14y ago
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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)
How do you factor 5x 30Y?
5x + 30Y = 5(x + 6Y)
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)
