Question 1127524
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;I will show you &nbsp;<U>MUCH &nbsp;FASTER &nbsp;way</U> &nbsp;to solve the system !



<pre>
The setup is still the same:


Let burgers be x, fries y, and drink z

x + y + z = 5.50       (1)
x + 0 + z = 4.00       (2)
x + y + 0 = 4.25       (3)



a)  To find "y", subtract eq(2) from eq(1). You will get

    y = 5.50 - 4.00 = 1.50.     The unknown "y" is just found !



b)  To find "z", subtract eq(3) from eq(1). You will get

    z = 5.50 - 4.25 = 1.25.     The unknown "z" is just found !



c)  To find "x", substitute the found values of y and z into eq(1) to get

    x + 1.50 + 1.25 = 5.50,   which implies

    x = 5.50 - 1.50 - 1.25 = 2.75.
</pre>

Just solved (!)


The structure of the system &nbsp;(1), &nbsp;(2), &nbsp;(3) &nbsp;is such that elimination works very straightforward for &nbsp;"y" &nbsp;and &nbsp;"z" &nbsp;(!).


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An educated student must see it after a quick look into the system.



An experienced student must recognize this way of solving immediately after reading the condition even without writing the system.



This problem is, &nbsp;actually, &nbsp;for &nbsp;<U>MENTAL</U>&nbsp; solution.