Note: I am assuming your second equation is:
x = 2y - 1
Answer:
The solution to the system of equations is:
Step-by-step explanation:
Given the system of equations
2x+y=3
x = 2y - 1
solving the system of equations using the elimination method
[tex]\begin{bmatrix}2x+y=3\\ x=2y-1\end{bmatrix}[/tex]
Arrange equation variables for elimination
[tex]\begin{bmatrix}2x+y=3\\ x-2y=-1\end{bmatrix}[/tex]
Multiply x-2y=-1 by2: 2x-4y=-2
[tex]\begin{bmatrix}2x+y=3\\ 2x-4y=-2\end{bmatrix}[/tex]
subtracting the equations
[tex]2x-4y=-2[/tex]
[tex]-[/tex]
[tex]\underline{2x+y=3}[/tex]
[tex]-5y=-5[/tex]
now solve -5y = -5 for y
[tex]-5y=-5[/tex]
Divide both sides by -5
[tex]\frac{-5y}{-5}=\frac{-5}{-5}[/tex]
Simplify
[tex]y=1[/tex]
For 2x+y=3 plug in y=1
[tex]2x+1=3[/tex]
subtract 1 from both sides
[tex]2x+1-1=3-1[/tex]
Simplify
[tex]2x=2[/tex]
Divide both sides by 2
[tex]\frac{2x}{2}=\frac{2}{2}[/tex]
Simplify
[tex]x=1[/tex]
Therefore, the solution to the system of equations is:
