Solve for x, y, z, a, b
b=1
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x=\frac{15}{49}\times \frac{7}{5}
Consider the first equation. Multiply both sides by \frac{7}{5}, the reciprocal of \frac{5}{7}.
x=\frac{3}{7}
Multiply \frac{15}{49} and \frac{7}{5} to get \frac{3}{7}.
x=\frac{3}{7} y=1 z=1 a=1 b=1
The system is now solved.
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