Solve the pair of equations simultaneously:
[tex]\begin{array}{l}
2 x-y=3 \\
x^2-x y=-4
\end{array}[/tex]
Answer (1)
Express y in terms of x from the first equation: y = 2 x − 3 .
Substitute this expression into the second equation: x 2 − x ( 2 x − 3 ) = − 4 .
Simplify and solve the resulting quadratic equation: x 2 − 3 x − 4 = 0 , which factors to ( x − 4 ) ( x + 1 ) = 0 , giving x = 4 or x = − 1 .
Find the corresponding y values: If x = 4 , y = 5 ; if x = − 1 , y = − 5 . The solutions are ( 4 , 5 ) , ( − 1 , − 5 ) .
Explanation
Problem Analysis We are given a system of two equations with two variables, x and y . Our goal is to find the values of x and y that satisfy both equations simultaneously. The equations are:
2 x − y = 3
x 2 − x y = − 4
Expressing y in terms of x From the first equation, we can express y in terms of x . This will allow us to substitute this expression into the second equation, resulting in a single equation with only x as a variable. We can then solve for x .
Isolating y From equation (1), 2 x − y = 3 , we can isolate y :
y = 2 x − 3
Substitution Now, we substitute this expression for y into equation (2), x 2 − x y = − 4 :
x 2 − x ( 2 x − 3 ) = − 4
Simplifying the equation Next, we simplify the equation: x 2 − 2 x 2 + 3 x = − 4 − x 2 + 3 x = − 4 x 2 − 3 x − 4 = 0
Solving for x Now we have a quadratic equation in terms of x . We can solve this by factoring: ( x − 4 ) ( x + 1 ) = 0 So, x = 4 or x = − 1
Solving for y Now we find the corresponding values of y for each value of x using the equation y = 2 x − 3 .
If x = 4 , then y = 2 ( 4 ) − 3 = 8 − 3 = 5 .
If x = − 1 , then y = 2 ( − 1 ) − 3 = − 2 − 3 = − 5 .
Solutions Therefore, the solutions are ( x , y ) = ( 4 , 5 ) and ( x , y ) = ( − 1 , − 5 ) .
Verification We can verify these solutions by substituting them back into the original equations.
For ( 4 , 5 ) :
2 ( 4 ) − 5 = 8 − 5 = 3 (Equation 1 is satisfied) ( 4 ) 2 − ( 4 ) ( 5 ) = 16 − 20 = − 4 (Equation 2 is satisfied)
For ( − 1 , − 5 ) :
2 ( − 1 ) − ( − 5 ) = − 2 + 5 = 3 (Equation 1 is satisfied) ( − 1 ) 2 − ( − 1 ) ( − 5 ) = 1 − 5 = − 4 (Equation 2 is satisfied)
Both solutions satisfy the original equations.
Final Answer The solutions to the system of equations are ( 4 , 5 ) and ( − 1 , − 5 ) .
Examples
Systems of equations are used in various real-world applications, such as determining the equilibrium point in economics where supply equals demand, designing structures in engineering where forces must balance, or planning flight paths in aviation to optimize fuel consumption and travel time. In this case, solving the system helps find the intersection points of a line and a hyperbola, which can represent different constraints or relationships in a model.
