Physics, 10.09.2019 17:10 xavierfox1721
In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a further step to isolate it. let's say we now have
3c+4d=5
2c+5d=2
none of these terms has a coefficient of 1. instead, we'll pick the variable with the smallest coefficient and isolate it. move the term with the lowest coefficient so that it's alone on one side of its equation, then divide by the coefficient. which of the following expressions would result from that process?
c=53β43d
c=1β52d
d=25β25c
d=54β34c
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part e - solving for two variables
now that you have one of the two variables in part d isolated, use substitution to solve for the two variables. you may want to review the multiplication and division of fractions and simplifying an expression primers.
enter the answer as two numbers (either fraction or decimal), separated by a comma, with c first.
need the answer with work shown for part e.
in advance
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In some cases, neither of the two equations in the system will contain a variable with a coefficient...
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