The points are: x = 0 and y = -1/10.
10y plus 40 equals 50y.
10y = 5 y = 5/10 Therefore, y = 0.5
10y = 2y - 6y + 7 10y =-4y + 7 10y +4y = 7 14y = 7 y = 1/2 = .5
(2, 1)
15y+14=2(5y+6) 15y+14=10y+12 15y+14-14=10y+12-14 15y=10y-2 15y-10y=10y-2-10y 5y=-2 y=-2/5
10y plus 40 equals 50y.
8x - 10y = 27x - 5y = 13First, we need to eliminate one of the variables. For this, we would like to have them with the same coefficient but different sign. We can do this if we multiply by -2 the second equation.8x - 10y = 2-2(7x - 5y) = -2(13)8x - 10y = 2-14x +10y = -26Now, we can add both equations, and solve for x.(8x - 10y) + (-14x + 10y) = (2) + (-26)(8x - 14x) +(-10y + 10y) = (2 - 26)-6x = - 24-6x/-6 = - 24/-6x = 4Substitute 4 for x in the first equation:8x - 10y = 28(4) - 10y = 232 - 10y = 232 - 32 - 10y = 2 - 32-10y = -30-10y/-10 = -30/-10y = 3Thus, the solution of the system is (4, 3).Check:
Treat it as a simultaneous equation question: -2x-10y = -12 3x+y = 4 Multiply all terms in the bottom equation by 10 in order to eliminate y: -2x-10y = -12 30x+10y = 40 Add both equations together: 28x+0 = 28 Divide both sides of the equation by 28 to find the value of x: x = 1 Substitute the value of x into the original equations to find the value of y: x = 1 and y = 1
10y = 5 y = 5/10 Therefore, y = 0.5
You're there. 4x = 10y - 8.[so subtract 10y from both sides, 4x - 10y = -8]multiply b.s. by -1 , gives -4x +10y = 8
10y = 2y - 6y + 7 10y =-4y + 7 10y +4y = 7 14y = 7 y = 1/2 = .5
10y = 9y + 4, so y = 4
x = 2 and y = 1 The lines intersect at the point (2, 1)
what is the solution of x-5y=10 and 2x-10y=20
1 If: 2x+5y = 16 and -5x-2y = 2 2 Then: 2*(2x+5y =16) and 5*(-5x-2y = 2) is equvalent to the above equations 3 Thus: 4x+10y = 32 and -25x-10y = 10 4 Adding both equations: -21x = 42 or x = -2 5 Solutions by substitution: x = -2 and y = 4
2
Write out equation - 10y - 5x= - 20 Negate equation 10y + 5x = 20 Isolate y variable 10y = - 5x + 20 Divide equation by 10 y= -1/2x + 2