Solve the following system using elimination:
{-2 x + 2 y + 3 z = 0 | (equation 1)
{-2 x - y + z = -3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)
Subtract equation 1 from equation 2:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x - 3 y - 2 z = -3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)
Multiply equation 2 by -1:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+3 y + 2 z = 3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)
Add equation 1 to equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+3 y + 2 z = 3 | (equation 2)
{0 x+5 y + 6 z = 5 | (equation 3)
Swap equation 2 with equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+3 y + 2 z = 3 | (equation 3)
Subtract 3/5 × (equation 2) from equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y - (8 z)/5 = 0 | (equation 3)
Multiply equation 3 by 5/8:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y - z = 0 | (equation 3)
Multiply equation 3 by -1:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)
Subtract 6 × (equation 3) from equation 2:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y+0 z = 5 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)
Divide equation 2 by 5:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)
Subtract 2 × (equation 2) from equation 1:
{-(2 x) + 0 y+3 z = -2 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
v0 x+0 y+z = 0 | (equation 3)
Subtract 3 × (equation 3) from equation 1:
{-(2 x)+0 y+0 z = -2 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)
Divide equation 1 by -2:
{x+0 y+0 z = 1 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)
Collect results:
Answer: {x = 1, y = 1, z = 0
Answer:
y - x = - 2 b
Step-by-step explanation:
Given-
3 x + 2 y = 5 a + b...................(i)
4 x - 3 y = a + 7 b.....................(ii)
<em><u>Multiply by 3 and 2 in equation (i) & (ii) respectively-</u></em>
<em><u>Add equation (iii) & (iv)-</u></em>
9 x + 6 y = 15 a + 3 a...................(iii)
<u>8 x - 6 y = 2 a + 14 b</u> ...................(iv)
17 x = 17 a + 17 b
<u>x = a + b</u>
Put x = a + b in equation (i)-
<em>2 y = 5 a + b - 3 a - 3 b</em>
<em>2 y = 2 a - 2 b</em>
<u>y = a - b </u>
<em>∴ y - x = a - b - ( a + b )</em>
<em>y - x = a - b - a - b</em>
y - x = - 2 b
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Answer:
Answer:
Distance is 13.6 units.
Which means it is 13
Step-by-step explanation:
Given a grid shows the positions of subway stop and house. The subway stop is located at (-7, 8) and house is located at (6, 4).
we have to find the distance between house and the subway stop.
Points are (-7, 8) and (6, 4).
Using distance formula,
D= /(x2-x1)^2+(y2-y1)^2} = (6-(-7))^2+(4-8)^2=(13)^2+16}={169+16}={185}=13.60147\sim13.6units.
Hence, distance between house and the subway stop is 13.6 units.
15:15
Which is simplified to
1:1
Hope this helps!