From the information available;
Let the numbers be x and y.
One number is 4 less than a second number. This would be tranlated as x is 4 less than y, or;
[tex]x=y-4[/tex]
Also, twice the second number is 30 more than 4 times the first, that is
[tex]\begin{gathered} 2\times y=30+(4\times x) \\ OR \\ 2y=30+4x \end{gathered}[/tex]
We now have a system of simultaneous equations which we shall solve as follows;
[tex]\begin{gathered} x=y-4---(1) \\ 2y=30+4x---(2) \\ \text{Substitute for x=y-4 into equation (2)} \\ 2y=30+4(y-4) \\ 2y=30+4y-16 \\ \text{Collect all like terms;} \\ 2y-4y=30-16 \\ -2y=14 \\ \text{Divide both sides by -2;} \\ -\frac{2y}{-2}=\frac{14}{-2} \\ y=-7 \\ \text{Substitute for the value of y into equation (1)} \\ x=y-4 \\ x=-7-4 \\ x=-11 \end{gathered}[/tex]
ANSWER:
The two numbers are;
[tex]-11\text{ and -7}[/tex]
