Algebra Quiz: Graphs As Sets Of Solutions

What this quiz covers

This quiz focuses on Graphs As Sets Of Solutions, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. No, because substitute: $3=2^2-1\rightarrow 3=5$ (false).
2. No, because substitute: $3=(2^2)-(-1)\rightarrow 3=5$ (true).
3. Yes, because substitute: $3=2^2-1\rightarrow 3=3$ (true). (correct answer)
4. Yes, because substitute: $2=3^2-1\rightarrow 2=8$ (true).

Explanation: This question tests your understanding of a fundamental idea: the graph of an equation in two variables is the set of all solution pairs (x, y) plotted on the coordinate plane—it's a visual representation of every pair that makes the equation true. To check if a point is on a graph, substitute its coordinates into the equation: if (a, b) is on the graph of y = x² - 1, then b must equal a² - 1. Substitute: does b = a² - 1? If yes, the point is on the graph (it's a solution). If no, it's not on the graph (not a solution). The graph and the equation are two ways of showing the same information! To verify if (2, 3) is on the graph of y = x² - 1: substitute x = 2 and y = 3 into the equation: 3 = (2)² - 1 = 4 - 1 = 3. Yes, 3 = 3, so the point is on the graph—it's a solution! Choice A correctly verifies the point is on the graph because it shows the accurate substitution 3 = 2² - 1 = 3, which is true. Choice B makes an arithmetic error when checking: it calculates 2² - 1 = 3 but then claims 3 = 5, which is incorrect—remember to compute exponents first! The point-on-graph test is super simple: take the point (a, b), substitute x = a and y = b into the equation, and see if you get a true statement. True = on the graph. False = not on the graph. That's it! Example: Is (4, 10) on it? Check: 10 = 3(4) - 1 = 11. False, so no!

Question 2

1. $(0,6)$, $(1,4)$, $(3,0)$ (correct answer)
2. $(0,6)$, $(1,5)$, $(3,1)$
3. $(0,4)$, $(1,2)$, $(3,0)$
4. $(0,6)$, $(2,2)$, $(4,0)$

Explanation: This question tests your understanding of a fundamental idea: the graph of an equation in two variables is the set of all solution pairs (x, y) plotted on the coordinate plane—it's a visual representation of every pair that makes the equation true. An equation like y = -2x + 6 (rewritten from 2x + y = 6) has infinitely many solutions—any (x, y) pair where y equals -2x + 6 works: (0, 6), (1, 4), (3, 0), and so on. Rather than listing them all (impossible!), we plot them all at once, and they form a line. Every point on that line represents a solution, and every solution to the equation appears as a point on the line. The graph IS the complete solution set! To find points on the graph of 2x + y = 6, we can choose x-values and solve for y (or vice versa): Choose x = 0: 2(0) + y = 6 → y = 6, giving point (0, 6). Choose x = 1: 2(1) + y = 6 → y = 4, giving point (1, 4). Choose x = 3: 2(3) + y = 6 → y = 0, giving point (3, 0). Each of these points is a solution, and plotting all such points (infinitely many!) creates the line. Choice A correctly identifies points on the graph because each satisfies the equation through accurate substitution and solving. Choice B makes an arithmetic error when checking: for (1,5), 2(1) + 5 = 7 ≠ 6, so it's not a solution—careful calculation is essential! For finding points on a graph: pick any x-value you want, substitute it into the equation, solve for y—that gives you a point (x, y) on the graph. Do this several times and you can plot points, then connect them to see the curve. Every (x, y) you find this way is a solution, and plotting all possible ones gives the complete graph!

Question 3

1. The single point $(2,1)$.
2. Only the points where $x=0$ or $y=0$.
3. All points $(x,y)$ such that $y=2x+1$. (correct answer)
4. All points $(x,y)$ such that $x=2y+1$.

Explanation: This question tests your understanding of a fundamental idea: the graph of an equation in two variables is the set of all solution pairs (x, y) plotted on the coordinate plane—it's a visual representation of every pair that makes the equation true. An equation like y = 2x + 1 has infinitely many solutions—any (x, y) pair where y equals 2x + 1 works: (0, 1), (1, 3), (2, 5), and so on. Rather than listing them all (impossible!), we plot them all at once, and they form a line. Every point on that line represents a solution, and every solution to the equation appears as a point on the line. The graph IS the complete solution set! The equation y = 2x + 1 has infinitely many solutions—for every x-value in the domain, there's a y-value making the equation true. Together, these (x, y) pairs form the solution set. When we graph them all, they create a straight line. This is why the graph is continuous: there's a solution for every x-value, and plotting them all gives us the continuous curve. Choice A correctly identifies the solution set because it describes all points (x, y) where y = 2x + 1, which is the complete set. Choice B confuses being a solution with being a specific type of point: every point on the graph is a solution, not just the intercepts. The whole line represents the solution set, not just certain highlighted points. Each point has equal status as a solution! Graph-equation relationship: if you have the graph, you can find the equation by analyzing features. If you have the equation, you can draw the graph by plotting solutions. They're two sides of the same coin—different ways to represent the same relationship between x and y. Being fluent in both is powerful!

Question 4

1. The graph is only the intercepts, because those are the easiest solutions to find.
2. The graph is the set of all ordered pairs $(x, y)$ that make $y = 2x + 1$ true. (correct answer)
3. The graph shows only integer solutions to $y = 2x + 1$.
4. The graph is a single point because there is one equation.

Explanation: This question tests your understanding of a fundamental idea: the graph of an equation in two variables is the set of all solution pairs (x, y) plotted on the coordinate plane—it's a visual representation of every pair that makes the equation true. An equation like y = 2x + 3 has infinitely many solutions—any (x, y) pair where y equals 2x + 3 works: (0, 3), (1, 5), (2, 7), and so on. Rather than listing them all (impossible!), we plot them all at once, and they form a line. Every point on that line represents a solution, and every solution to the equation appears as a point on the line. The graph IS the complete solution set! The equation y = 2x + 1 has infinitely many solutions—for every x-value in the domain, there's a y-value making the equation true. Together, these (x, y) pairs form the solution set. When we graph them all, they create a straight line. This is why the graph is continuous: there's a solution for every x-value, and plotting them all gives us the continuous curve. Choice B correctly explains that the graph shows all solutions because the graph is the set of all ordered pairs (x, y) that make y = 2x + 1 true, which is the complete solution set forming the line. Choice C confuses being a solution with being a specific type of point: every point on the graph is a solution, not just integer solutions. The whole line represents the solution set, not just certain highlighted points. Each point has equal status as a solution! Graph-equation relationship: if you have the graph, you can find the equation by analyzing features. If you have the equation, you can draw the graph by plotting solutions. They're two sides of the same coin—different ways to represent the same relationship between x and y. Being fluent in both is powerful!

Question 5

1. The graph is the set of all points $(x,y)$ such that $x^2+y^2=25$. (correct answer)
2. The graph shows only the integer solutions to $x^2+y^2=25$.
3. The graph is a line because the equation has two variables.
4. The graph is the set of all points where $x^2+y^2$ is less than $25$.

Explanation: This question tests your understanding of a fundamental idea: the graph of an equation in two variables is the set of all solution pairs (x, y) plotted on the coordinate plane—it's a visual representation of every pair that makes the equation true. Different equations create different graphs because they have different solution sets: linear equations like y = mx + b have solutions forming a straight line, quadratic equations like y = x² have solutions forming a parabola, and circle equations like x² + y² = 25 have solutions forming a circle. The equation determines which points are solutions, and plotting all those points creates the characteristic shape! The equation x² + y² = 25 has infinitely many solutions—for every x-value in the domain, there's a y-value making the equation true. Together, these (x, y) pairs form the solution set. When we graph them all, they create a circle. This is why the graph is continuous: there's a solution for every valid x-value, and plotting them all gives us the continuous curve. Choice A correctly explains that the graph shows all solutions because it describes the graph as the set of all points satisfying the equation, which forms the circle. Choice B thinks the graph has only finitely many solutions, but equations in two variables typically have infinitely many solutions: every point on the circle is a solution, not just integer ones—the graph shows them all as a continuous curve! Different equations, different graphs: linear equations (y = mx + b) give straight lines because their solutions lie on a line. Quadratic equations (y = x²) give parabolas because their solutions trace a U-shape. Circle equations (x² + y² = r²) give circles because solutions are all at distance r from the origin. The equation type determines the graph shape!

Question 6

1. Yes, because substitute: $2(3)+2=8\rightarrow 8=8$ (true). (correct answer)
2. No, because substitute: $2(3)+2=8\rightarrow 6=8$ (false).
3. Yes, because substitute: $2(2)+3=8\rightarrow 7=8$ (true).
4. No, because substitute: $2(3)+2=8\rightarrow 10=8$ (false).

Explanation: This question tests your understanding of a fundamental idea: the graph of an equation in two variables is the set of all solution pairs (x, y) plotted on the coordinate plane—it's a visual representation of every pair that makes the equation true. To check if a point is on a graph, substitute its coordinates into the equation: if (a, b) is on the graph of 2x + y = 8, then b must equal 8 - 2a, or directly plug in to see if 2a + b = 8 holds true. To verify if (3, 2) is on the graph of 2x + y = 8: substitute x = 3 and y = 2 into the equation: 2(3) + 2 = 6 + 2 = 8, which equals 8. Yes, 8 = 8, so the point is on the graph—it's a solution! Choice A correctly verifies the point is on the graph because it shows the accurate substitution 2(3) + 2 = 8 leading to 8 = 8, which is true. Choice B makes an arithmetic error when checking: it calculates 2(3) + 2 as 6 + 2 = 8 but then claims 6 = 8, which is incorrect—careful addition is key here! The point-on-graph test is super simple: take the point (a, b), substitute x = a and y = b into the equation, and see if you get a true statement. True = on the graph. False = not on the graph. That's it! Example: Is (2, 5) on y = 3x - 1? Check: 5 = 3(2) - 1 = 5. True, so yes!

Question 7

1. No, because substitute: $4=|-1|+3\rightarrow 4=-1+3$ (false).
2. Yes, because substitute: $4=|-1|+3\rightarrow 4=1+3$ (true). (correct answer)
3. No, because substitute: $-1=|4|+3\rightarrow -1=7$ (false).
4. Yes, because substitute: $4=|1|+3\rightarrow 4=5$ (true).

Explanation: This question tests your understanding of a fundamental idea: the graph of an equation in two variables is the set of all solution pairs (x, y) plotted on the coordinate plane—it's a visual representation of every pair that makes the equation true. To check if a point is on a graph, substitute its coordinates into the equation: if (a, b) is on the graph of y = |x| + 3, then b must equal |a| + 3. Substitute: does b = |a| + 3? If yes, the point is on the graph (it's a solution). If no, it's not on the graph (not a solution). The graph and the equation are two ways of showing the same information! To verify if (-1, 4) is on the graph of y = |x| + 3: substitute x = -1 and y = 4 into the equation: 4 = |-1| + 3 = 1 + 3 = 4. Yes, 4 = 4, so the point is on the graph—it's a solution! Choice B correctly verifies the point is on the graph because it shows the accurate substitution 4 = |-1| + 3 = 4, which is true. Choice A makes an arithmetic error when checking: it forgets the absolute value and calculates |-1| as -1 instead of 1. When verifying points on graphs, careful arithmetic is crucial—one small mistake makes a solution look like a non-solution or vice versa! The point-on-graph test is super simple: take the point (a, b), substitute x = a and y = b into the equation, and see if you get a true statement. True = on the graph. False = not on the graph. That's it!

Question 8

1. Every point on the graph satisfies the equation, but some solutions to the equation are not on the graph
2. Every solution to the equation corresponds to a point on the graph, but some points on the graph don't satisfy the equation
3. The graph represents all solutions to the equation, and every point on the graph satisfies the equation (correct answer)
4. The graph shows only the positive solutions to the equation, while negative solutions exist but aren't displayed

Explanation: The graph of an equation in two variables is precisely the set of all solutions plotted in the coordinate plane. Every point (x,y) on the graph satisfies the equation, and every solution (x,y) to the equation appears as a point on the graph. This is the fundamental definition of what it means to graph an equation.

Question 9

1. Calculate $(-2)^2 + 3$ and compare the result to 7 to see if they're equal (correct answer)
2. Check whether $x = -2$ gives a positive y-value, since parabolas open upward
3. Graph the equation and visually estimate whether $(-2, 7)$ appears to be on the curve
4. Use the quadratic formula to solve $x^2 + 3 = 7$ and check if $x = -2$ is a solution

Explanation: When you need to determine whether a specific point lies on the graph of an equation, you're testing whether that point satisfies the equation. A point $(x, y)$ lies on a graph if and only if substituting the x-coordinate into the equation produces the given y-coordinate. To check if $(-2, 7)$ lies on $y = x^2 + 3$, substitute $x = -2$ into the equation: $y = (-2)^2 + 3 = 4 + 3 = 7$. Since this gives us $y = 7$, which matches the y-coordinate of our point, $(-2, 7)$ does lie on the graph. This confirms that choice A is the correct approach. Choice B is flawed because knowing that a parabola opens upward doesn't tell us anything specific about whether a particular point lies on it. Yes, $x = -2$ gives a positive y-value, but that's not sufficient to determine if the point is on the curve. Choice C relies on visual estimation, which is unreliable and unnecessary when you can calculate the exact answer. Graphing might seem reasonable, but it introduces potential error and is less precise than algebraic verification. Choice D overcomplimplicates the problem. While solving $x^2 + 3 = 7$ would give you $x = \pm 2$, this backward approach is unnecessarily complex when you can directly substitute the given x-value. Remember: to verify if a point lies on a graph, always substitute the x-coordinate into the equation and check if you get the corresponding y-coordinate. This direct substitution method is foolproof and efficient.

Question 10

1. The point is not a solution because circles cannot have negative coordinates
2. The point is a solution because $3^2 + (-4)^2 = 9 + 16 = 25$ (correct answer)
3. The point is not a solution because it doesn't lie on the positive portion of the circle
4. The point is a solution only if we consider the absolute values: $|3|^2 + |-4|^2 = 25$

Explanation: To determine if (3, -4) is on the graph, we substitute into the equation: 3² + (-4)² = 9 + 16 = 25. Since this equals 25, the point satisfies the equation and therefore lies on the graph. The sign of coordinates is irrelevant - what matters is whether the coordinates satisfy the equation.

Question 11

1. The point appears because when $x = 5$, we get $y = |5 - 2| = |3| = 3$ (correct answer)
2. The point appears because 5 and 3 are both positive numbers, matching the V-shape
3. The point appears because it's located 3 units from the vertex of the V-shape
4. The point appears because the distance formula gives $\sqrt{(5-2)^2 + (3-0)^2} = \sqrt{18}$

Explanation: When you encounter absolute value functions, remember that to verify if a point lies on the graph, you substitute the x-coordinate into the equation and check if it produces the given y-coordinate. For the equation $y = |x - 2|$, let's test whether $(5, 3)$ is actually on the graph. Substituting $x = 5$: $y = |5 - 2| = |3| = 3$. Since this gives us the point $(5, 3)$, the point is indeed on the graph. This direct substitution method is the fundamental way to verify any point on any function. Looking at the wrong answers: Choice B incorrectly assumes that having positive coordinates automatically means a point fits the V-shape, but this ignores the specific equation entirely—many positive coordinate points don't lie on this particular graph. Choice C mentions being "3 units from the vertex," but this is imprecise and doesn't demonstrate understanding of how absolute value functions work. The vertex is at $(2, 0)$, and while there are distance relationships, that's not how we verify points on graphs. Choice D applies the distance formula between $(5, 3)$ and $(2, 0)$, which calculates the distance between two points but has nothing to do with whether a point satisfies the equation. Choice A demonstrates the correct mathematical reasoning by using direct substitution to verify the relationship between x and y coordinates. Remember: To check if any point lies on a graph, always substitute the x-value into the equation and see if you get the corresponding y-value. This works for linear, quadratic, absolute value, and all other types of functions.

Question 12

1. Whether $x = 4$ when $y = 0$ in the original equation
2. Whether $y = 5$ when $x = 0$ in the original equation
3. Whether substituting $x = 4$ and $y = 5$ makes the equation true (correct answer)
4. Whether the point $(4, 5)$ satisfies $y = 2x + 13$ instead

Explanation: A point lies on the graph of an equation if and only if the coordinates of that point satisfy the equation. To check if (4,5) is on the graph of 2x + y = 13, substitute x = 4 and y = 5: 2(4) + 5 = 8 + 5 = 13 ✓. Since this makes the equation true, the point is on the graph.

Question 13

1. $(0, -1)$, $(1, 0)$, and $(2, 3)$ because they form a nice pattern
2. $(0, -1)$, $(1, 1)$, and $(-1, 1)$ because they include positive and negative values
3. $(1, 2)$, $(2, 5)$, and $(3, 10)$ because they show the quadratic growth pattern
4. $(-1, 0)$, $(0, -1)$, and $(1, 0)$ because they satisfy the equation when substituted (correct answer)

Explanation: When you're asked to find points on a graph, you need to verify that each point actually satisfies the given equation. The only way to know for certain that a point lies on the graph of $y = x^2 - 1$ is to substitute the x-coordinate into the equation and check if you get the corresponding y-coordinate. Let's test option D by substituting each point into $y = x^2 - 1$:

• For $(-1, 0)$: $y = (-1)^2 - 1 = 1 - 1 = 0$ ✓
• For $(0, -1)$: $y = (0)^2 - 1 = 0 - 1 = -1$ ✓
• For $(1, 0)$: $y = (1)^2 - 1 = 1 - 1 = 0$ ✓

Question 14

1. Only points on the boundary line $y = 2x + 1$ represent solutions to the inequality
2. Points in the shaded region and on the boundary line both represent solutions to the inequality (correct answer)
3. Points in the shaded region represent solutions, but points on the boundary line do not
4. The graph shows some solutions, but infinitely many solutions exist outside the shaded region

Explanation: For the inequality y ≤ 2x + 1, the solution set includes all points where y equals 2x + 1 (the boundary line) AND all points where y is less than 2x + 1 (the shaded region below the line). The graph represents the complete solution set - every solution appears on the graph, and every point on the graph is a solution.

Question 15

1. Only $(2,5)$ and $(2,4)$
2. All four points
3. Only $(1,2)$ and $(0,-1)$
4. $(1,2)$, $(0,-1)$, and $(2,5)$ (correct answer)

Explanation: This question tests your understanding of a fundamental idea: the graph of an equation in two variables is the set of all solution pairs (x, y) plotted on the coordinate plane—it's a visual representation of every pair that makes the equation true. To check if a point is on a graph, substitute its coordinates into the equation: if (a, b) is on the graph of y = 3x - 1, then b must equal 3a - 1. Substitute: does b = 3a - 1? If yes, the point is on the graph (it's a solution). If no, it's not on the graph (not a solution). The graph and the equation are two ways of showing the same information! To verify if the points are on the graph of y = 3x - 1: for (1, 2), 2 = 3(1) - 1 = 2 (yes); for (0, -1), -1 = 3(0) - 1 = -1 (yes); for (2, 5), 5 = 3(2) - 1 = 5 (yes); for (2, 4), 4 = 3(2) - 1 = 5 (no, 4 ≠ 5). So three points are solutions! Choice C correctly identifies which points satisfy the equation because it lists exactly the ones where substitution gives a true statement: (1,2), (0,-1), and (2,5). Choice D correctly identifies some solutions but claims a non-solution is on the graph: checking (2, 4) in y = 3x - 1: 4 = 6 - 1 = 5, but 4 ≠ 5. Just because some points work doesn't mean all points work—each point must be verified individually, or we trust the graph shows exactly the solution set. The point-on-graph test is super simple: take the point (a, b), substitute x = a and y = b into the equation, and see if you get a true statement. True = on the graph. False = not on the graph. That's it!

Question 16

1. All ordered pairs $(x, y)$ such that $y = x^2 - 4$. (correct answer)
2. Only the intercepts of the parabola.
3. Exactly two solutions because the equation is quadratic.
4. A single $x$-value where $y$ is always 0.

Explanation: This question tests your understanding of a fundamental idea: the graph of an equation in two variables is the set of all solution pairs (x, y) plotted on the coordinate plane—it's a visual representation of every pair that makes the equation true. Different equations create different graphs because they have different solution sets: linear equations like y = mx + b have solutions forming a straight line, quadratic equations like y = x² have solutions forming a parabola, and circle equations like x² + y² = 25 have solutions forming a circle. The equation determines which points are solutions, and plotting all those points creates the characteristic shape! The equation y = x² - 4 has infinitely many solutions—for every x-value in the domain, there's a y-value making the equation true. Together, these (x, y) pairs form the solution set. When we graph them all, they create a parabola. This is why the graph is continuous: there's a solution for every x-value, and plotting them all gives us the continuous curve. Choice B correctly explains that the graph shows all solutions because it identifies the graph as all ordered pairs (x, y) such that y = x² - 4, which is the complete solution set. Choice D thinks the graph has only finitely many solutions, but equations in two variables typically have infinitely many solutions: every point on the parabola is a solution. For y = x² - 4, there are infinitely many (x, y) pairs that work—the graph shows them all as a continuous curve! Understanding graphs as solution sets helps you see why we graph equations: rather than making huge tables of x and y values, we can SEE all solutions at once. The graph of y = x² shows every single pair where the y-value is the square of the x-value—infinitely many pairs—all in one picture. That's the power of graphing! Different equations, different graphs: linear equations (y = mx + b) give straight lines because their solutions lie on a line. Quadratic equations (y = x²) give parabolas because their solutions trace a U-shape. Circle equations (x² + y² = r²) give circles because solutions are all at distance r from the origin. The equation type determines the graph shape!

Question 17

1. Yes. Substitute: $2 = |-1| + 1 \rightarrow 2 = 1 + 1$ (true). (correct answer)
2. No. Substitute: $2 = |-1| + 1 \rightarrow 2 = -1 + 1$ (false).
3. No. Substitute: $2 = |-1| + 1 \rightarrow 2 = 1 - 1$ (false).
4. Yes. Substitute: $-1 = |2| + 1 \rightarrow -1 = 3$ (true).

Explanation: This question tests your understanding of a fundamental idea: the graph of an equation in two variables is the set of all solution pairs (x, y) plotted on the coordinate plane—it's a visual representation of every pair that makes the equation true. To check if a point is on a graph, substitute its coordinates into the equation: if (a, b) is on the graph of y = x² - 5, then b must equal a² - 5. Substitute: does b = a² - 5? If yes, the point is on the graph (it's a solution). If no, it's not on the graph (not a solution). The graph and the equation are two ways of showing the same information! To verify if (-1, 2) is on the graph of y = |x| + 1: substitute x = -1 and y = 2 into the equation: 2 = |-1| + 1. Simplifying: 2 = 1 + 1, so 2 = 2. Yes, 2 = 2, so the point IS on the graph—it's a solution! Choice A correctly verifies the point is on the graph because it shows the accurate substitution 2 = |-1| + 1 leading to 2 = 1 + 1 = 2, which is true. Choice B makes an arithmetic error when checking: it calculates |-1| + 1 but claims 2 = -1 + 1 = 0 instead of 1 + 1 = 2. When verifying points on graphs, careful arithmetic is crucial—one small mistake makes a solution look like a non-solution or vice versa! The point-on-graph test is super simple: take the point (a, b), substitute x = a and y = b into the equation, and see if you get a true statement. True = on the graph. False = not on the graph. That's it! Example: Is (2, 5) on y = 3x - 1? Check: 5 = 3(2) - 1 = 5. True, so yes! Is (4, 10) on it? Check: 10 = 3(4) - 1 = 11. False, so no! This works for any equation and any point.

Question 18

1. No. Substitute: $2(3)+2=8 \rightarrow 8=10$ (false).
2. No. Substitute: $2(3)+2=8 \rightarrow 6=8$ (false).
3. Yes. Substitute: $2(3)+2=8 \rightarrow 8=8$ (true). (correct answer)
4. Yes. Substitute: $2(2)+3=8 \rightarrow 7=8$ (true).

Explanation: This question tests your understanding of a fundamental idea: the graph of an equation in two variables is the set of all solution pairs (x, y) plotted on the coordinate plane—it's a visual representation of every pair that makes the equation true. To check if a point is on a graph, substitute its coordinates into the equation: if (a, b) is on the graph of y = x² - 5, then b must equal a² - 5. Substitute: does b = a² - 5? If yes, the point is on the graph (it's a solution). If no, it's not on the graph (not a solution). The graph and the equation are two ways of showing the same information! To verify if (3, 2) is on the graph of 2x + y = 8: substitute x = 3 and y = 2 into the equation: 2(3) + 2 = 8. Simplifying: 6 + 2 = 8, so 8 = 8. Yes, 8 = 8, so the point IS on the graph—it's a solution! Choice A correctly verifies the point is on the graph because it shows the accurate substitution 2(3) + 2 = 8 leading to 8 = 8, which is true. Choice B makes an arithmetic error when checking: it calculates 2(3) + 2 = 8 but then claims 6 = 8 instead of 8 = 8. When verifying points on graphs, careful arithmetic is crucial—one small mistake makes a solution look like a non-solution or vice versa! The point-on-graph test is super simple: take the point (a, b), substitute x = a and y = b into the equation, and see if you get a true statement. True = on the graph. False = not on the graph. That's it! Example: Is (2, 5) on y = 3x - 1? Check: 5 = 3(2) - 1 = 5. True, so yes! Is (4, 10) on it? Check: 10 = 3(4) - 1 = 11. False, so no! This works for any equation and any point.

Question 19

1. No. Substitute: $1 = 2^2 - 3 \rightarrow 1 = 7$ (false).
2. Yes. Substitute: $2 = 1^2 - 3 \rightarrow 2 = -2$ (true).
3. Yes. Substitute: $1 = 2^2 - 3 \rightarrow 1 = 1$ (true). (correct answer)
4. No. Substitute: $1 = 2^2 - 3 \rightarrow 1 = -1$ (false).

Explanation: This question tests your understanding of a fundamental idea: the graph of an equation in two variables is the set of all solution pairs (x, y) plotted on the coordinate plane—it's a visual representation of every pair that makes the equation true. To check if a point is on a graph, substitute its coordinates into the equation: if (a, b) is on the graph of y = x² - 5, then b must equal a² - 5. Substitute: does b = a² - 5? If yes, the point is on the graph (it's a solution). If no, it's not on the graph (not a solution). The graph and the equation are two ways of showing the same information! To verify if (2, 1) is on the graph of y = x² - 3: substitute x = 2 and y = 1 into the equation: 1 = 2² - 3. Simplifying: 1 = 4 - 3, so 1 = 1. Yes, 1 = 1, so the point IS on the graph—it's a solution! Choice A correctly verifies the point is on the graph because it shows the accurate substitution 1 = 2² - 3 leading to 1 = 4 - 3 = 1, which is true. Choice B makes an arithmetic error when checking: it calculates 2² - 3 = 4 - 3 = 1 but then claims 1 = 7 instead of 1 = 1. When verifying points on graphs, careful arithmetic is crucial—one small mistake makes a solution look like a non-solution or vice versa! The point-on-graph test is super simple: take the point (a, b), substitute x = a and y = b into the equation, and see if you get a true statement. True = on the graph. False = not on the graph. That's it! Example: Is (2, 5) on y = 3x - 1? Check: 5 = 3(2) - 1 = 5. True, so yes! Is (4, 10) on it? Check: 10 = 3(4) - 1 = 11. False, so no! This works for any equation and any point.

Question 20

1. No. Substitute: $4+0=4 \rightarrow 5=4$ (false).
2. Yes. Substitute: $0+4=4 \rightarrow 0=4$ (true).
3. No. Substitute: $4+0=4 \rightarrow 4=0$ (false).
4. Yes. Substitute: $4+0=4 \rightarrow 4=4$ (true). (correct answer)

Explanation: This question tests your understanding of a fundamental idea: the graph of an equation in two variables is the set of all solution pairs (x, y) plotted on the coordinate plane—it's a visual representation of every pair that makes the equation true. To check if a point is on a graph, substitute its coordinates into the equation: if (a, b) is on the graph of y = x² - 5, then b must equal a² - 5. Substitute: does b = a² - 5? If yes, the point is on the graph (it's a solution). If no, it's not on the graph (not a solution). The graph and the equation are two ways of showing the same information! To verify if (4, 0) is on the graph of x + y = 4: substitute x = 4 and y = 0 into the equation: 4 + 0 = 4. Simplifying: 4 = 4. Yes, 4 = 4, so the point IS on the graph—it's a solution! Choice B correctly verifies the point is on the graph because it shows the accurate substitution 4 + 0 = 4 leading to 4 = 4, which is true. Choice A makes an arithmetic error when checking: it calculates 4 + 0 = 4 but then claims 4 = 0 instead of 4 = 4. When verifying points on graphs, careful arithmetic is crucial—one small mistake makes a solution look like a non-solution or vice versa! The point-on-graph test is super simple: take the point (a, b), substitute x = a and y = b into the equation, and see if you get a true statement. True = on the graph. False = not on the graph. That's it! Example: Is (2, 5) on y = 3x - 1? Check: 5 = 3(2) - 1 = 5. True, so yes! Is (4, 10) on it? Check: 10 = 3(4) - 1 = 11. False, so no! This works for any equation and any point.