23 = 7x + 7x - 5 = 14x - 5 14x = 23 + 5 = 28 x = 28/14 = 2
This equation is unsolvable without knowing one of the values of either x or y.
75
22+22+28+28=100.
it equals 28
23 = 7x + 7x - 5 = 14x - 5 14x = 23 + 5 = 28 x = 28/14 = 2
28
Set up the equation and solve for z: 28 + z = 56 (next, subtract 28 from each side of the equaition to solve) z = 28
a2+30a+56=0 , solve for a Using the quadratic formula, you will find that: a=-2 , a=-28
n = 28
x + 4y = -14 eqn12x + 3y = 13 eqn2Using the elimination method we multiply eqn1 by 22x + 8y = -28 eqn1bSubtract eqn2 from eqn1b2x + 8y = -28 eqn1b2x + 3y = 13 eqn25y =-41y = -8 and 1/5substituting this into eqn1 we getx +4 (-41/5) = -14x = (14*5) / (41 *4)x = (35/82)
This equation is unsolvable without knowing one of the values of either x or y.
75
22+22+28+28=100.
it equals 28
it is 28
3x + 7 = -4x + 28 implies 7x + 7 = 28