Explanation
We are given the following:
[tex]\begin{gathered} 4x-3y=10\text{ \lparen equation 1\rparen} \\ 4x-2y=3\text{ \lparen equation 2\rparen} \end{gathered}[/tex]
We are required to solve the simultaneous equation given using substitution method.
This can be achieved as follows:
[tex]\begin{gathered} \text{ From equation 1, } \\ 4x-3y=10 \\ 4x=10+3y\text{ \lparen equation 3\rparen} \\ \\ \text{ Substitute for ''4x'' in equation 2.} \\ 4x-2y=3 \\ (10+3y)-2y=3 \\ 10+3y-2y=3 \\ 3y-2y=3-10 \\ y=-7 \end{gathered}[/tex]
Now, we get the value of x as follows:
[tex]\begin{gathered} \text{ From equation 3} \\ 4x=10+3y \\ where \\ y=-7 \\ 4x=10+3(-7) \\ 4x=10-21 \\ 4x=-11 \\ x=\frac{-11}{4} \end{gathered}[/tex]
Hence, the answer is:
[tex]x=\frac{-11}{4};y=-7[/tex]
