30a+28b=7
21a+21b=5
find a and b
You asked:
Given \( 30 a + 28 b = 7 \), \( 21 a + 21 b = 5 \), evaluate \( a \), \( b \).
MathBot Answer:
The system of linear equations has exactly one solution.
\[a = \frac{1}{6}\]
\[b = \frac{1}{14}\]
Solve \(30 a + 28 b = 7\) for \(a\). \[a = \frac{7}{30} - \frac{14 b}{15}\]
Substitute \(\frac{7}{30} - \frac{14 b}{15}\) for \(a\) in \(21 a + 21 b = 5\) and simplify. $$\begin{aligned}21 a + 21 b &= 5 \\ 21 \left(\frac{7}{30} - \frac{14 b}{15}\right) + 21 b &= 5 \\ b &= \frac{1}{14} \end{aligned}$$
Use substitution of the numerical value of \(b\) to get the values of \(a\). $$\begin{aligned}a &= \frac{7}{30} - \frac{14 b}{15} \\ a &= - \frac{1 \cdot 14}{14 \cdot 15} + \frac{7}{30} \\ a &= \frac{1}{6}\end{aligned}$$