Question

If \(x+y=32,\) and \(x-y=12,\) then \(y=?\) F. \(\quad 6\) G. 10 H. 20 J. 22 K. 44

Short answer

\(x+y=32\) \(x-y=12\) Answer: G. 10

Step-by-step solution

Rewrite the given equations

We have the system of equations: \(x+y=32 \hspace{1cm}(1)\) \(x-y=12 \hspace{1cm}(2)\)

Add equations (1) and (2) together to eliminate one of the variables

We want to eliminate one of the variables so that we can solve the system of equations. In this case, we will eliminate y by adding both equations together: \((x+y)+(x-y)=32+12\)

Simplify and solve for x

Simplify the equation and solve for x: \(2x+0=44\) \(2x=44\) \(x=22\)

Substitute the value of x into either equation (1) or (2) to solve for y

Substitute the value of x into equation (1): \(22+y=32\)

Solve for y

Solve the equation for y: \(y=32-22\) \(y=10\) The value of y is 10, which corresponds to answer choice G.

Key concepts

System of Equations

Understanding how to solve a system of equations is a fundamental skill in mathematics, especially on tests like the ACT. A system of equations is simply a set of two or more equations that have the same variables. The solution to this system is the set of values that makes all the equations true simultaneously.

For example, if we have two equations, one being \(x + y = 32\) and the other \(x - y = 12\), the solution to this system would be the values of \(x\) and \(y\) that make both of these equations valid at the same time. To solve this, we use various methods such as graphing, substitution, elimination, or algebraic manipulation. The goal is to find the point where the equations 'meet' or intersect, which represents the common solution to both equations.

In the ACT Math problem provided, we're dealing with a system of linear equations, which are straight-line graphs. This means their point of intersection can be found algebraically, owing to the simplicity of dealing with straight lines.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and transforming equations to make them easier to solve. It involves a variety of operations, including addition, subtraction, multiplication, division, and factoring.

In step 2 of our solution, we added the two equations to eliminate \(y\), which illustrates the importance of algebraic manipulation in solving systems of equations. Simplifying complex or unwieldy expressions down to their most basic form is essential, as it makes the path to the solution clearer and more straightforward. Effective algebraic manipulation often requires a thoughtful approach to understanding how one operation will affect others and planning several steps ahead.

This skill not only serves to solve typical algebra problems but also forms the basis for more advanced topics in mathematics, where a strong grasp on manipulating expressions is crucial.

Equation Solving Techniques

Several tried-and-true equation solving techniques can be employed when faced with a system of equations. As seen in our ACT Math problem, direct addition or subtraction of equations is a common method to eliminate a variable. This is part of a broader strategy known as the elimination method. Once one variable is eliminated, it becomes a simpler matter to solve for the remaining variable.

Another technique is substitution, where one equation is solved in terms of one variable and that expression is substituted into the other equation. This method is highly effective when dealing with equations that do not easily lend themselves to addition or subtraction.

Additionally, if the equations are particularly complicated, factoring or using the quadratic formula may be necessary. Understanding which technique to apply and when can make the difference in solving problems quickly and correctly, which is especially important in a timed exam like the ACT.