Is the relationship between the values in the table a direct variation, an inverse variation, or neither
1 answer:
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Answer:
1) isolate x for -x-5y-5z=2: x = -2 - 5y - 5z
substitute x = -2 -5y - 5z
{4(-2 - 5y - 5z) - 5y + 4z = 19}
{-2 - 5y - 5z + 5y - z = -20}
simplify
{-25y - 16z - 8 = 19}
{-6z - 2 = -20}
isolate z for -6z - 2 = -20: z = 3
substitute z = 3
{-25y - 16 * 3 - 8 = 19}
simplify
{-25y - 56 = 19}
isolate y for -25y - 56 = 19: y = -3
for x = -2 - 5y - 5z
substitute z = 3, y = -3
x = -2 - 5(-3) - 5 * 3
-2 - 5(-3) - 5 * 3 = -2
x = -2
x = -2, z = 3, y = -3
2)isolate x for -4x - 5y - z = 18: x = -(18+5y+z)/4
substitute x = -(18+5y+z)/4
{-2(-(18+5y+z)/4) - 5y - 2z = 12}
{-2(-(18+5y+z)/4) + 5y + 2z = 4}
simplify
{(-5y-3z+18)/2 = 12}
{(15y+5z+18)/2 = 4}
isolate y for (-5y-3z+18)/2 = 12: y = -(3z+6)/5
substitute y = -(3z+6)/5
{15((-3z+6)/5)+5z+18/2 = 4}
simplify
{-2z=4}
isolate z for -2z=4: z = -2
for y = - 3z+6/5
substitute z = -2
y = - 3(-2)+6/5
- 3(-2)+6/5 = 0
y = 0
for x = - 18+5y+z/4
substitute z = -2, y = 0
x = - 18+5*0-2/4
- 18+5*0-2/4 = -4
x = -4
x = -4, z = -2, y = 0
3) x = -1, z = -4, y = -4
4) x = 4, z = 0, y = 2
5) r = 1, t = 1, s = 3
6) x = 0, z = -3, y = 0
<u><em>work for 3, 4, 5, and 6 is below</em></u>
