\left\{ \begin{array} { l } { b = 1 } \\ { 1.5 k + b = 15 } \end{array} \right.
Solve for b, k
b=1
k = \frac{28}{3} = 9\frac{1}{3} \approx 9.333333333
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1.5k+1=15
Consider the second equation. Insert the known values of variables into the equation.
1.5k=15-1
Subtract 1 from both sides.
1.5k=14
Subtract 1 from 15 to get 14.
k=\frac{14}{1.5}
Divide both sides by 1.5.
k=\frac{140}{15}
Expand \frac{14}{1.5} by multiplying both numerator and the denominator by 10.
k=\frac{28}{3}
Reduce the fraction \frac{140}{15} to lowest terms by extracting and canceling out 5.
b=1 k=\frac{28}{3}
The system is now solved.
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