Question

Solve the system by the method of substitution. $$\left\\{\begin{array}{l}2 x-y+2=0 \\ 4 x+y-5=0\end{array}\right.$$

Short answer

The solution to the system of equations is x = 0.5, y = 2.

Step-by-step solution

Solve First Equation for Y

The first goal is to isolate 'y' in the first equation. By adding 'y' and subtracting '2' from both sides of the 1st equation \(2x - y + 2 = 0\), then the equation becomes \(y = 2x + 2\).

Substitute Y into the Second Equation

Now we substitute \(y = 2x + 2\) into the second equation \(4x + y -5 = 0\), which becomes \(4x + (2x + 2) - 5 =0\). Simplifying this equation gives \(6x - 3 = 0\).

Solve for X

By solving \(6x - 3 = 0\) for 'x', we add '3' to both sides and divide by '6'. This results in \(x = 3/6 = 0.5\).

Substitute X into the first equation

Now we substitue the value of 'x' obtained earlier into the first equation \(y = 2x + 2\). This gives \(y = 2(0.5) + 2 = 2\).

Interpret the Solution

This means that 'x = 0.5' and 'y = 2' is the solution to the system of equations. Representing the point (0.5, 2) in a Cartesian plane, it is clearly the intersection of the lines represented by the original pair of equations.

Key concepts

Substitution Method

The substitution method is a powerful algebraic approach for solving systems of linear equations. It involves replacing one variable with an equivalent expression, thereby reducing the system to a single equation with one variable.

For instance, in the example provided, we tackle a pair of linear equations. The first step is to express 'y' in terms of 'x' using the first equation. Once 'y' is isolated and rewritten as an expression containing 'x', you substitute this expression into the second equation. This substitution effectively eliminates 'y', allowing you to solve for 'x'.

After solving for 'x', you substitute this value back into the expression for 'y'. The beauty of this method is its simplicity and systematic approach that can be applied to a wide range of linear systems. Through step-by-step substitution, you transform a system of equations into a more manageable form, leading to the solution.

System of Linear Equations

A system of linear equations consists of two or more equations with the same variables. The goal is to find values for the variables that satisfy all equations simultaneously.

In our exercise, the system is made up of two equations in two variables, 'x' and 'y'. The system can represent geometrical lines on a Cartesian plane, and the solution to the system corresponds to the point(s) where the lines intersect.

Sometimes, systems of linear equations can have one solution (intersecting lines), no solution (parallel lines), or infinitely many solutions (coincident lines). The method you choose to solve the system, whether it be substitution, elimination, or graphical analysis, depends on the specifics of the problem and your proficiency with each technique.

Algebraic Method

Algebraic methods are a range of techniques that use algebra to solve equations and systems. These include substitution, elimination, and graphing methods, among others.

The substitution method, as showcased in the exercise, is an algebraic technique where we solve one of the equations for one variable and then use this solution to simplify the other equation(s). Algebraic methods are grounded in logical steps and arithmetic operations, making sure that what we perform on one side of an equation we also do on the other side to maintain equality.

Each step needs to be clear and justifiable, and understanding the underlying principles of algebra—like performing inverse operations to isolate variables—is crucial in mastering these methods for solving systems of equations.

Solving for a Variable

Solving for a variable means rearranging an equation to isolate the variable on one side of the equation while placing all other terms on the opposite side. This process often involves several steps of adding, subtracting, multiplying, or dividing terms on both sides of the equation.

In our exercise, we solved the first equation for 'y' and then proceeded to find 'x' in the reduced form of the second equation. It's essential to perform each operation carefully to avoid errors. After finding the value for one variable, you substitute it back into one of the original equations to solve for the other variable. The ability to solve for a particular variable is fundamental in algebra and is a technique that will be used in more complex mathematics and applications beyond.