Here's the question you clicked on:
vegas14
If you were solving a system of equations and you came to a statement like 1 = 3, what do you know about the solution(s) to this system? Part 2: Solve the following system and show all of your work. x − 2y = 14 x + 3y = 9
it tells you that there is no solution i.e. if graphing the 2 equations, the lines do NOT cross. ...
x- 2y = 14 x+3y = 9 need a common factor to cancel x to to solve for y though the cancelling of terms . -1 is good. So I'll use it x-2y = 14 (-1) x +3y = 9(-1) - Multiplying a negative number to one of the equation to both sides of it to change a variable from a negative value to a positive one or vise verse to cancel one of the terms to solve for the other though the addition of terms x -2y = 14 -x-3y= -9 ---------------- - Multiplication and addition of terms 0x -5y = 5 -5y = 5 - Addition 1/-5* -5y = 5 * 1/-5 - Multiplying the reciprocal of a coefficient to both sides of the equation to solve for y y = 5/-5 - Multiplication y = -1 - Division let solve for x x-2y = 14 - one of the original equations x - 2(-1) = 14 - substituting y with -1 x + 2 = 14 - Multiplication x + 2 -2 = 14- 2 - Adding the additive inverse of a number to both sides of the equation to move it to the other side of it x= 12 - Addition proof: x- 2y = 14 x+3y = 9 - original equation 12-2(-1) = 14 12+ 3(-1) = 9 - substituting y and x with -1, and 12 12+ 2 = 14 12-3 = 9 - Multiplication 14 = 14 9 = 9 - Addition ...
i think it helps...