Given the equations:
[tex]\begin{gathered} 1)y=2x-5 \\ 2)2x-5y=1 \end{gathered}[/tex]To solve this equation system, you have to replace the first one into the second one:
[tex]2x-5(2x-5)=1[/tex]Solve the term in parentheses using the distributive propperty of multiplication
[tex]\begin{gathered} 2x-5\cdot2x-5\cdot(-5)=1 \\ 2x-10x+10=1 \end{gathered}[/tex]Solve for x
[tex]\begin{gathered} 2x-10x+25=1 \\ -8x=1-25 \\ -8x=-24 \\ x=-\frac{24}{-8} \\ x=3 \end{gathered}[/tex]Finally replace the calculated value of x in the first equation and solve for y
[tex]\begin{gathered} y=2\cdot3-5 \\ y=1 \end{gathered}[/tex]Using the second equation you can prove if the calculations are correct:
[tex]\begin{gathered} 2x-5y=1 \\ \text{for x=3 and y=1 } \\ 2\cdot3-5\cdot1=6-5=1 \end{gathered}[/tex]The calculations check, the values are x=3 and y=1
