Linear equations

In algebra, an equivalence transformation is a change you make to an equation that produces a new equation with the exact same solution set as before. If the new equation ends up with extra solutions or missing solutions, then the transformation is not equivalent.

To check whether squaring really keeps solutions unchanged, start with a simple linear equation:

$x=5$

This equation has only one solution, so its solution set is: $\{5\}$

Now square the left and right sides:

$(x{)}^2=(5{)}^2$ $x^2=25$

Notice that the equation is no longer linear; it has become quadratic after squaring.

Next, solve $x^2=25$. Taking square roots means considering both signs:

$x=\pm \sqrt{25}$ $x=5\text{ or }x=-5$

So the new solution set is: $\{5,-5\}$

Compare the two sets:

• Original set: $\{5\}$
• After squaring: $\{5,-5\}$

Because an additional solution appeared ($x=-5$), squaring did not keep the solution set exactly the same. That extra value is an extraneous solution, so the statement must be False.

Since squaring both sides can create solutions that were not in the original equation, it is not an equivalence transformation.

In mathematics, an equivalence transformation is an operation that turns one equation into another while keeping the exact same solution set. A useful way to judge this is to ask whether the operation is reversible.

If you can go from Equation A to Equation B and then undo the step to return from B to A without introducing or losing any possibilities, then the transformation is equivalent.

Check what happens with a straightforward equation: $x=3$ Its solution set is $\{3\}$. Squaring both sides gives: $x^2=9$ But reversing by taking the square root does not return a single value: it gives $3$ or $-3$, so the solution set becomes $\{3,-3\}$.

Squaring is not truly reversible because different inputs can lead to the same output (it is many-to-one). For example: $\begin{array}{rl}(2{)}^2& =4\\ (-2{)}^2& =4\end{array}$ When you square, you erase the sign information. That missing information is why squaring can produce extraneous solutions—values that satisfy the squared equation but not the original.

Because squaring can expand the solution set by adding extra values, it fails the requirement for an equivalence transformation.

Since squaring is not reversible in a way that preserves the exact solution set, the statement is False.

We are given the linear equation $4x+7=x-2$. The task is to determine the value of $x$ by getting $x$ alone on one side. This kind of rearranging is a key skill in Coordinate Geometry and Linear Modelling.

Collect the terms with $x$ on the left by subtracting $x$ from both sides:

$\begin{array}{rl}4x-x+7& =x-x-2\\ 3x+7& =-2\end{array}$

Now remove the $+7$ from the left side by subtracting $7$ from both sides:

$\begin{array}{rl}3x+7-7& =-2-7\\ 3x& =-9\end{array}$

Since $3x=-9$, divide both sides by $3$ to solve:

$\begin{array}{rl}\frac{3x}{3}& =\frac{-9}{3}\\ x& =-3\end{array}$

Compare our solution $x=-3$ to the choices.

• Option A is $6$, which is not correct because it is positive and not equal to our result.
• Options B and C are decimals and also do not equal $-3$.

Our solved value is exactly Option D.

To solve this linear equation using the Substitution of Options approach, we substitute each proposed value of $x$ into both sides. The correct choice makes the Left-Hand Side (LHS) equal the Right-Hand Side (RHS).

The equation is: $4x+7=x-2$

Substitute $x=6$:

LHS: $4(6)+7=24+7=31$ RHS: $6-2=4$

Since $31\ne 4$, Option A is incorrect.

Substitute $x=-2.33$:

LHS: $4(-2.33)+7=-9.32+7=-2.32$ RHS: $-2.33-2=-4.33$

Since $-2.32\ne -4.33$, Option B is incorrect.

Substitute $x=0.75$:

LHS: $4(0.75)+7=3+7=10$ RHS: $0.75-2=-1.25$

Since $10\ne -1.25$, Option C is incorrect.

Substitute $x=-3$:

LHS: $4(-3)+7=-12+7=-5$ RHS: $-3-2=-5$

Both sides are equal, so $x=-3$ is the solution.

To solve $4x+7=x-2$ with a graphing calculator, treat the two sides as separate linear functions:

$y_1=4x+7$ $y_2=x-2$

The solution is the $x$-value where the two graphs intersect.

Enter $y_1$ and $y_2$ into your GDC, plot both lines, then use the intersection tool. The calculator will display the intersection point $(x,y)$.

The graphs meet at $(-3,-5)$.

Because we are solving for $x$, we take the $x$-coordinate: $x=-3$

Match the graphing result to the options, and note why others fail if substituted:

• If $x=6$: $4(6)+7=31$ and $6-2=4$. Since $31\ne 4$, this is incorrect.
• If $x=-2.33$: $4(-2.33)+7=-2.32$ and $-2.33-2=-4.33$. This is incorrect.
• If $x=0.75$: $4(0.75)+7=10$ and $0.75-2=-1.25$. This is incorrect.

The intersection gives $x=-3$, which corresponds to option D.

Checking directly: $4(-3)+7=-12+7=-5$ $-3-2=-5$ Both sides match, so the solution is confirmed.

In algebra, the word equivalent has a precise meaning. Using the Direct Definition, two equations are equivalent exactly when they have the identical solution set.

In other words:

1. Any value that solves the first equation also solves the second.
2. Any value that solves the second equation also solves the first.

This is a key idea from Number and Algebra Foundations that helps ensure we don’t change the solutions while rewriting equations.

The statement asks whether it is true that equations are called equivalent when they share the same solution set.

For example, look at these two equations: $x+2=5$ $2x=6$

Each equation has the solution set $\{3\}$. Since the solutions are exactly the same, these two equations fit the definition of being equivalent. So the statement lines up perfectly with the standard definition used in algebra.

Now check the answer choices to confirm which one is correct:

• Option B is incorrect because the statement is actually the accepted definition in mathematics.
• Option C is incorrect because equivalence is not restricted to linear equations. For instance, $x^2=1$ and $(x-1)(x+1)=0$ are both quadratic and are equivalent because they share the same solution set $\{1,-1\}$.
• Option D is incorrect because equations don’t need to have the same degree to be equivalent. For example, $x=2$ (degree 1) and $x^2-4x+4=0$ (degree 2) both have the solution set $\{2\}$, so they are equivalent.

Because equivalence depends only on having the same solution set (and not on degree or type), the statement is True.

In mathematics, especially in Number and Algebra Foundations, “equivalent equations” means the equations describe the same solutions. The question is whether this depends only on the solution set, or whether there must also be rules about the equation being linear or having a matching degree.

Option C says equations are only equivalent if they are linear (degree 1).

A counter-example shows this is not true. Consider:

1. $x^2=9$
2. $|x|=3$

Both have the solution set $\{-3,3\}$. Since the solutions match exactly, the equations are equivalent even though neither equation is linear. So Option C is incorrect.

Option D says equations have to be the same degree to be equivalent.

Here is a counter-example:

1. $x-5=0$ (Degree 1)
2. $(x-5{)}^2=0$ (Degree 2)

Both equations lead to the solution $x=5$, so each has the solution set $(\{5\})$. Since the solution sets match, the equations are equivalent even though their degrees are different. Therefore, Option D is also false.

After removing the extra restrictions in Options C and D, and since Option B (False) goes against the basic definition, the correct choice is Option A.

Two equations are equivalent if and only if they produce the same solution set (each equation has exactly the solutions of the other). That is the standard MYP definition.

To decide if an equation is linear, start by checking where the variable sits. The equation is:

$\frac{4}{x}+2=6$

The key part is the fraction $\frac{4}{x}$. In Coordinate Geometry and Linear Modelling, linear equations match the pattern $ax+b=c$, where the variable is in the numerator and has power 1, not placed underneath a fraction line.

Using index rules from Number and Algebra Foundations, the term $\frac{4}{x}$ can be expressed with a negative exponent:

$\frac{4}{x}=4x^{-1}$

A linear equation requires the variable to be in the form $x^1$. Since the exponent here is $-1$, the equation is not linear; it fits the category of a rational equation instead.

Option A and Option B both say the equation is linear, but their reasons don’t hold up. Having only one variable (Option A) is not enough—equations like $x^2=4$ also have one variable but are not linear. Also, the variable is not used as a linear term (Option B) because it is in the denominator rather than appearing as $x$ in a polynomial term.

Now check the claim that there are no real solutions. Solve it:

$\begin{array}{rl}\frac{4}{x}+2& =6\\ \frac{4}{x}& =4\\ x& =1\end{array}$

Because $x=1$ is a real solution and it works in the original equation, the “no real solutions” statement is false.

After ruling out the incorrect statements, the correct conclusion is that the equation is not linear. The variable appears in the denominator, which makes the equation rational, not linear.

To decide if $\frac{4}{x}+2=6$ is linear, focus on the placement of $x$. Linear equations have variables in the numerator with exponent 1, like $ax+b=c$.

Here, $x$ is located in the denominator in $\frac{4}{x}$. That feature signals a rational equation, not a linear one.

To evaluate one option, solve for $x$:

$\begin{array}{rl}\frac{4}{x}+2& =6\\ \frac{4}{x}& =4\\ 4& =4x\\ x& =1\end{array}$

Since $x=1$ is a valid real value, we can eliminate the statement that the equation has no real solutions.

The idea that an equation is linear just because it contains one variable is incorrect. For instance, $x^2=9$ and $\sqrt{x}=5$ each involve only $x$ but are not linear. Because our equation’s structure is not linear, this reasoning should be eliminated.

Even though $x$ may look like it is to the power of 1, it is in the denominator. Using exponent laws:

$\frac{4}{x}=4x^{-1}$

A linear equation requires the variable’s exponent to be exactly $1$. Here it is $-1$, so the equation is not linear.

With the other options eliminated, the correct conclusion is that the equation is not linear. Because $x$ is in the denominator, it is a rational equation, matching the reasoning in Option C.