Solve the following system:
{t = 2 h + 92 | (equation 1)
t = 3 h + 77 | (equation 2)
Express the system in standard form:
{t - 2 h = 92 | (equation 1)
t - 3 h = 77 | (equation 2)
Subtract equation 1 from equation 2:
{t - 2 h = 92 | (equation 1)
0 t - h = -15 | (equation 2)
Multiply equation 2 by -1:
{t - 2 h = 92 | (equation 1)
0 t+h = 15 | (equation 2)
Add 2 × (equation 2) to equation 1:
{t+0 h = 122 | (equation 1)
0 t+h = 15 | (equation 2)
Collect results:
Answer: |t = 122 and h=15
+245 Source: http://www.coolmath.com/algebra/12-2x2-systems-of-equations/03-solving-by-elimination-addition-01
t = 77 + 3h
t = 92 + 2h
the point of elimination is to eliminate one of the variables in the system of equations.
Lets eliminate t
t = 77 + 3h
- 1(t = 92 + 2h) multiply by -1
t= 77 + 3h now add the two equations
+ -t= -92 - 2h
0 = -15 +h
h = 15
Now plug in 15 for h in order to solve for t.
t = 77 + 3 (15) = 77 + 45 = 122
So then now we know t =122 and h =15
