Use the substitution method to solve the system
$\left\{\begin{array}{l}
-x+y=0 \\
4 x-3 y=-2
\end{array}\right.$
One solution: $\square$
No solution
Answer (1)
Solve the first equation for y : y = x .
Substitute y into the second equation: 4 x − 3 x = − 2 .
Solve for x : x = − 2 .
Substitute x back into the equation for y : y = − 2 . The solution is ( − 2 , − 2 ) .
Explanation
Analyze the problem We are given the system of equations: − x + y = 0 4 x − 3 y = − 2 We will solve this system using the substitution method.
Solve for y From the first equation, we can express y in terms of x :
y = x
Substitute into second equation Now, substitute this expression for y into the second equation: 4 x − 3 ( x ) = − 2
Solve for x Simplify and solve for x :
4 x − 3 x = − 2 x = − 2
Solve for y Substitute the value of x back into the equation y = x to find the value of y :
y = − 2
Verify the solution So the solution is x = − 2 and y = − 2 . Let's check this solution in both original equations: Equation 1: − ( − 2 ) + ( − 2 ) = 2 − 2 = 0 (Correct) Equation 2: 4 ( − 2 ) − 3 ( − 2 ) = − 8 + 6 = − 2 (Correct) Thus, the solution is x = − 2 and y = − 2 .
State the solution The solution to the system of equations is x = − 2 and y = − 2 , which can be written as the ordered pair ( − 2 , − 2 ) .
Examples
Systems of equations are used in various real-world applications, such as determining the break-even point for a business, calculating the optimal mix of ingredients in a recipe, or modeling traffic flow in a city. In this case, the system of equations could represent the relationship between the cost and revenue of a product, where x and y represent the number of units sold and the profit, respectively. Solving the system helps determine the point at which the cost equals the revenue, indicating the break-even point.
