Question 1

Solve the system of equation by the elimination method {1/3x+1/2y=1/2{1/6x-1/3y=5/6(x,y)=(_, _)

Solve the system of equation by the elimination method {1/3x+1/2y=1/2{1/6x-1/3y=5/6(x-example-1

Question 2

Solution

- The solution steps to solve the system of equations by elimination is given below:

$\begin{gathered} (x)/(3)+(y)/(2)=(1)/(2)\text{ \lparen Equation 1\rparen} \\ \\ (x)/(6)-(y)/(3)=(5)/(6)\text{ \lparen Equation 2\rparen} \\ \\ \text{ Multiply Equation 2 by 2} \\ 2*((x)/(6)-(y)/(3))=(5)/(6)*2 \\ \\ (x)/(3)-(2y)/(3)=(5)/(3)\text{ \lparen Equation 3\rparen} \\ \\ \\ \text{ Now, }(x)/(3)\text{ is common to both Equations 1 and 3.} \\ \\ \text{ We can therefore subtract both equations to eliminate }x. \\ \text{ We have:} \\ \text{ Equation 1 }-\text{ Equation 3} \\ \\ (x)/(3)+(y)/(2)-((x)/(3)-(2y)/(3))=(1)/(2)-(5)/(3) \\ \\ (x)/(3)-(x)/(3)+(y)/(2)+(2y)/(3)=(1)/(2)-(5)/(3)=(3)/(6)-(10)/(6) \\ \\ (y)/(2)+(2y)/(3)=-(7)/(6) \\ \\ (3y)/(6)+(4y)/(6)=-(7)/(6) \\ \\ (7y)/(6)=-(7)/(6) \\ \\ \therefore y=-1 \\ \\ \text{ Substitute the value of }y\text{ into any of the equations, we have:} \\ (1)/(3)x+(1)/(2)y=(1)/(2) \\ (1)/(3)x+(1)/(2)(-1)=(1)/(2) \\ \\ (1)/(3)x=(1)/(2)+(1)/(2) \\ \\ (1)/(3)x=1 \\ \\ \therefore x=3 \end{gathered}$

Final Answer

The answer is:

$\begin{gathered} x=3,y=-1 \\ \\ \therefore(x,y)=(3,-1) \end{gathered}$

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