Problem 36 - Algebra 2

Question

The coordinates \((x, y)\) of a point in a plane are the solution of the system \(\left\\{\begin{array}{l}{2 x+3 y=13} \\ {5 x+7 y=31}\end{array} . \text { Find the coordinates of the point. }\right.\)

Step-by-Step Solution

Answer

The coordinates of the point are \(x, y) = (2, 3)\).

Step 1: Write Down the Equations

Start by clearly writing out the system of equations provided:1. \(2x + 3y = 13\)2. \(5x + 7y = 31\)

Step 2: Choose a Method for Solving

Decide on a method to solve the system, such as substitution or elimination. For this system, the elimination method is suitable.

Step 3: Manipulate Equations to Eliminate a Variable

To eliminate one variable, manipulate the equations by multiplying them with suitable numbers to get equal coefficients for either of the variables. For instance, multiply the first equation by 5 and the second equation by 2:1. \(5(2x + 3y) = 5(13)\) resulting in \(10x + 15y = 65\)2. \(2(5x + 7y) = 2(31)\) resulting in \(10x + 14y = 62\)

Step 4: Subtract Equations to Solve for One Variable

Subtract the second modified equation from the first to solve for one variable:\((10x + 15y) - (10x + 14y) = 65 - 62\)The result is \(y = 3\).

Step 5: Substitute the Known Value to Solve for the Other Variable

Substitute \(y = 3\) back into one of the original equations to solve for \(x\). Using the first equation \(2x + 3y = 13\):\(2x + 3(3) = 13\)\(2x + 9 = 13\)\(2x = 4\)\(x = 2\).

Key Concepts

Elimination Method

One efficient technique for solving a system of linear equations, like the one involving the coordinates \( (x, y) \) of a point, is the elimination method. This method focuses on eliminating one variable to easily solve for the other. By either adding or subtracting the equations from one another, we aim to cancel out one of the variables.
To eliminate a variable in our example, we look for coefficients for \( x \) or \( y \) that are opposites or the same in both equations. Here, we saw that multiplying the first equation by 5 and the second by 2 resulted in both equations having the term \( 10x \) which then can be eliminated by subtraction.

• Multiply each equation by an appropriate value that will result in a common coefficient for one of the variables.
• Add or subtract the equations to cancel out one variable and solve for the other.

This method is especially useful when the coefficients of both equations are such that one can be made the negative of the other through multiplication, allowing for easy elimination.

Substitution Method

The substitution method is another pathway to finding the solution for a system of equations. It is particularly handy when one of the equations can be solved easily for one variable; that variable is then substituted into the other equation.
Using our initial system of equations, \(2x + 3y = 13\) and \(5x + 7y = 31\), if we could isolate \(x\) or \(y\) in one equation, we could substitute that expression in the other equation to find the second variable.

• Solve one of the equations for one variable.
• Substitute this expression into the other equation to find the second variable.

For our problem, the elimination method was more suitable due to the coefficients involved, but substitution could still have been used if we had identified one equation that was easily solvable for either \(x\) or \(y\).

Algebraic Manipulation

Algebraic manipulation is the use of algebraic methods to rewrite and solve equations in different forms, which is integral in both the elimination and substitution methods. Mastering algebraic manipulation is crucial to solving systems of equations effectively.
In the context of the presented problem, algebraic manipulation involved multiplying equations to form equivalent systems and adding or subtracting equations to eliminate variables. It can also involve dividing or factoring when necessary.

• Use operations such as multiplication, division, addition, and subtraction to modify equations.
• Employ these skills in combination with the substitution or elimination method for solving the system.

Good algebraic manipulation allows for a smoother transition between steps and often simplifies the process of finding the solution to a system of equations.