#1**+1 **

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**

Guest Aug 3, 2017

#2**+1 **

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.

TheXSquaredFactor Aug 3, 2017

#3**+2 **

Here's one more method :

Using X^{2}'s nomenclature

If product of [A_{1} * B_{2} ] - [ A_{2} * B_{1} ] 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

CPhill Aug 3, 2017

#5**0 **

It's just the intersection of two lines. Since they only intersect at one point,there is only one solution.

frasinscotland Aug 3, 2017

#6**+1 **

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.

TheXSquaredFactor
Aug 4, 2017