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

Sagot :

taskmasters
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)



24demarsc

Answer:

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.)

Step-by-step explanation:

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