Write a system of equations with a solution of (4, -3).

A linear equation can be written in several forms. "Standard Form" is #ax+by=c# where #a#, #b# and #c# are constants (numbers).
We want to make two equations that
(i) have this form,
(ii) do not have all the same solutions (the equations are not equivalent), and
(iii) #(4, -3)# is a solution to both.
#ax+by=c#. We want #a#, #b# and #c# so that
#a(4)+b(-3)=c# (This will make (i) and (iii) true.) ;

Answered by 24demarsc | 2024-06-12

1. So we have the system of equation. (4, -3)
let’s create another system of equation.
Let a = 4
and Let b = -3
First equation
=> Let’s pick any numbers to be used. Let’s have number 5
=> 5a + 5b
=> 5 (4) + 5 (-3)
=> 20 + (-15)
negative and positive is equals to negative
=> 20 – 15
=> 5 (this the first value of our equation, let this be the value of X)
Second
=> Pick another number, Let’s have 6
=> 6a + 6b
=> 6 (4) + 6 (-3)
=> 24 + (-18)
=> 24 – 18
=> 6 (this is the value of our second equation, let this value be Y)
Now, we have (A, B) = (4, -3) and (X, Y) = (5,6)

Answered by taskmasters | 2024-06-24

To create a system of equations with a solution of (4, -3), we can use the equations: 2 x + 3 y = − 1 and x − y = 7 . Substituting the point (4, -3) into both equations confirms it is a solution.
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Answered by taskmasters | 2024-10-05