Solve the following system:
{2 x + 3 y = 20 | (equation 1)
6 x - y = 20 | (equation 2)
Swap equation 1 with equation 2:
{6 x - y = 20 | (equation 1)
2 x + 3 y = 20 | (equation 2)
Subtract 1/3 × (equation 1) from equation 2:
{6 x - y = 20 | (equation 1)
0 x+(10 y)/3 = 40/3 | (equation 2)
Multiply equation 2 by 3/10:
{6 x - y = 20 | (equation 1)
0 x+y = 4 | (equation 2)
Add equation 2 to equation 1:
{6 x+0 y = 24 | (equation 1)
0 x+y = 4 | (equation 2)
Divide equation 1 by 6:
{x+0 y = 4 | (equation 1)
0 x+y = 4 | (equation 2)
Collect results:
Answer: | x = 4 and y=4
The question does not ask for the solution of the system of equations but rather how many solutions exist. There are only 3 possibilities. I will list them for you.
1. No solution
2. One solution
3. Infinitely many solutions
Currently, both equations are written in the form of \(Ax+By=C\). However, in a system, the system looks like the following:
{ \(A_1x+B_1y=C_1\)
{\(A_2x+B_2y=C_2\)
Now, let's compare the ratios of the system
{\(2x+3y=20\)
{\(6x-y=20\)
When I say compare the ratio, I mean that you want to look at the relationship of \(\frac{A_1}{A_2}\) and \(\frac{B_1}{B_2}\) and \(\frac{C_1}{C_2}\).
Let's look at them:
\(\frac{A_1}{A_2}=\frac{2}{6}=\frac{1}{3}\)
\(\frac{B_1}{B_2}=\frac{3}{-1}=-3\)
\(\frac{C_1}{C_2}=\frac{20}{20}=1\)
Since \(\frac{A_1}{A_2}\neq\frac{B_1}{B_2}\neq\frac{C_1}{C_2}\), there is only one solution.
Here's one more method :
Using X2's nomenclature
If product of [A1 * B2 ] - [ A2 * B1 ] is not 0 then we will always have one solution...so...
[2 * -1 ] - [6 * 3 ] = -2 - 18 = -20 .....so....we will have only one solution
It's just the intersection of two lines. Since they only intersect at one point,there is only one solution.
To say that 2 nonlinear lines intersects at a common point is correct. However, it is not given info that they are indeed nonlinear; you must prove it.