Is this a system of equations?
2X + 5Y = 4
X + 5Y = 7
solve for X using bottom equation
X = 5Y - 7
insert into top equation
2(5Y - 7) + 5Y = 4
distribute
10Y - 14 + 5Y = 4
solve for Y
15Y = 18
Y = 18/15 = 6/5
use bottom equation to find X
X +5(6/5) = 7
X = 1
check with bottom equation
1 + 5(6/5) = 7
7 = 7
not true for top equation
( sorry, I forgot what happens then-resubmit question )
8x + 20y
y=x+5
2X + 5Y = 205Y = - 2X + 20Y = (- 2/5)X + 4================now, reading off the function,Y intercept = 4===========
(4, -1)
3x + 4y - 5z and -2x - 5y + 4z Added together, give (3-2)x + (4-5)y + (-5+4)z = x - y - z
-2x = 20 + 5y-2x - 20 = 5y-2/5x - 4 = yX Intercept = (-10,0) or -10Y Intercept = (0,-4) or -4
8x + 20y
y=x+5
2X + 5Y = 205Y = - 2X + 20Y = (- 2/5)X + 4================now, reading off the function,Y intercept = 4===========
(4, -1)
2x+5y=20 Solve for y -2x -2x 5y=20-2x 5 5 5 y=4-2x 5
3x + 4y - 5z and -2x - 5y + 4z Added together, give (3-2)x + (4-5)y + (-5+4)z = x - y - z
x = 4 and y = -1 (4. -1)
2X+5Y+30=0 So you subtract 30 from each side. 2X+5Y+30-30=0-30. So you know 2X+5Y=-30. now there are many different possibilities for X and Y unless there is a specification. here are some possibilities=X=0,Y=6 or X=A5,Y=0 or X=5,Y=4
2.6
2x + 7 + 5x - 4 - x = (2x + 5x - x) + (7 - 4) = 6x + 3
If 2x+5y = 4 and y^2 = x+4 then by combining the equations into a single quadratic equation and solving it the points of contact are made at (12, -4) and (-7/2, 3/2)