The sum of the measures of the angles of a triangle is 180. The sum of the measures of the second and third angles is five times the measure of the first angle. The third angle is 14 more than the second. Let x, y, and z represent the measures of the first, second, and third angles, respectively. Find the measures of the three angles.
Solve the following system:
{y + z = 5 x | (equation 1)
z = y + 14 | (equation 2)
x + y + z = 180 | (equation 3)
Express the system in standard form:
{-(5 x) + y + z = 0 | (equation 1)
0 x - y + z = 14 | (equation 2)
x + y + z = 180 | (equation 3)
Add 1/5 × (equation 1) to equation 3:
{-(5 x) + y + z = 0 | (equation 1)
0 x - y + z = 14 | (equation 2)
0 x+(6 y)/5 + (6 z)/5 = 180 | (equation 3)
Multiply equation 3 by 5/6:
{-(5 x) + y + z = 0 | (equation 1)
0 x - y + z = 14 | (equation 2)
0 x+y + z = 150 | (equation 3)
Add equation 2 to equation 3:
{-(5 x) + y + z = 0 | (equation 1)
0 x - y + z = 14 | (equation 2)
0 x+0 y+2 z = 164 | (equation 3)
Divide equation 3 by 2:
{-(5 x) + y + z = 0 | (equation 1)
0 x - y + z = 14 | (equation 2)
0 x+0 y+z = 82 | (equation 3)
Subtract equation 3 from equation 2:
{-(5 x) + y + z = 0 | (equation 1)
0 x - y+0 z = -68 | (equation 2)
0 x+0 y+z = 82 | (equation 3)
Multiply equation 2 by -1:
{-(5 x) + y + z = 0 | (equation 1)
0 x+y+0 z = 68 | (equation 2)
0 x+0 y+z = 82 | (equation 3)
Subtract equation 2 from equation 1:
{-(5 x) + 0 y+z = -68 | (equation 1)
0 x+y+0 z = 68 | (equation 2)
0 x+0 y+z = 82 | (equation 3)
Subtract equation 3 from equation 1:
{-(5 x)+0 y+0 z = -150 | (equation 1)
0 x+y+0 z = 68 | (equation 2)
0 x+0 y+z = 82 | (equation 3)
Divide equation 1 by -5:
{x+0 y+0 z = 30 | (equation 1)
0 x+y+0 z = 68 | (equation 2)
0 x+0 y+z = 82 | (equation 3)
x = 30, y = 68, z = 82
