Linear and quadratic systems of equations on the SAT test
SAT subscore: Passport to advanced mathematics
Studying linear and quadratic systems of equations
On the SAT test linear and quadratic systems of equations are part of passport to advanced mathematics subscore that includes 9 advanced topics (see the full topics list on the top menu).
Linear and quadratic systems of equations topic is the third topic of passport to advances mathematics subscore. Start learning passport to advances mathematics subscore with its first topic (quadratic equations and quadratic functions) and continue to its second topic (graphing quadratic functions).
Linear and quadratic systems of equations topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.
Finish studying heart of algebra subscore topics before you study this topic or any other passport to advances mathematics subscore topic. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding passport to advanced mathematics subscore topics). Particularly before learning this topic learn heart of algebra linear equations topic and systems of linear equations topic.
Linear and quadratic systems of equations- summary
Linear and quadratic systems of equations include 2 equations: a linear equation and a quadratic equation:
A linear equation y=mx+b is an algebraic equation in which each term has an exponent of one. It is called linear because it can be graphed as a straight line in the xy plane.
A quadratic equation y=ax2+bx+c is an equation that has a squared term (a variable multiplied by itself) and is graphed as a parabola in the xy plane. Linear and quadratic system can be solved algebraically or graphically.
Graphical solution of linear and quadratic systems of equations is at the intersection point of the line and the parabola. This intersection can happen 0, 1 or 2 times.
Algebraic solution of linear and quadratic systems of equations can be done by substituting an expression for a variable (plugging an expression instead of a variable). This will leave us with one equation with one variable that we can solve.
Graphical solution of linear and quadratic systems of equations
To solve a system of equations graphically, graph both equations in the same coordinate system. The solutions to the system are in the points where the two graphs intersect.
Graphical solution of linear and quadratic systems of equations steps
Step 1: Graph the linear equation as a straight line in the xy plane:
To draw the graph, mark 2 points and draw a line between them. To make sure you have found the right points it is best to mark a third point. If there is no mistake all the points will be on the line. Which points to choose? Mark the points of intersection with the x-axis (y=0) and intersection with the y-axis (x=0) and then select a third point that is located between these two points.
For more details about graphing a line learn the graphic presentation of a linear function topic.
Step 2: Graph the quadratic equation as a parabola in the xy plane:
When we graph a quadratic function, we need to cover the vertex, the decreasing part and the increasing part. Plug some x values into the function to calculate their corresponding y values, plot the (x,y) coordinates in the xy plan and sketch a parabola through all the points.
For more details about graphing a parabola learn the graphing quadratic functions topic.
Step 3: See the coordinates of the intersection points of the line and the parabola:
The intersection can happen 0, 1 or 2 times, therefore a linear and quadratic system of equations can have 0, 1 or 2 solutions.
The graphs below show these 3 different possibilities of the solutions.
The position of the linear equation relative to the vertex of the parabola
Given the linear equation in a form of y=b (with no slope), the number of the solutions can be determined using the vertex of the parabola.
For more details about graphing the vertex of a parabola learn the vertex point of a parabola topic.
The different locations possibilities and the resulting numbers of solution are listed below.
If the vertex point of a parabola (minimum or maximum) is equal to b (b=y of the vertex), then there is 1 intersection point between the parabola and the line (at the vertex), therefore there is 1 solution to the equation system.
Consider the following example:
How many intersection points has the system of equations y=(x+1)2-4 and y=-4?
The vertex point of the parabola is (-1,-4), there is only one vertex point on the parabola. The equation of the line is y=-4, therefore the function will intersect the parabola at the vertex point and there is 1 solution to the equations system.
See a graphical presentation after the next paragraphs.
If the parabola has a minimum point and b is below the vertex (b<y of the vertex) there are no intersection points between the parabola and the line, therefore there are no solutions to the systems of the equation.
Consider the following example:
How many intersection points has the system of equations y=(x+1)2-4 and y=-7?
The vertex point of the parabola is (-1,-4), this is a minimum point since any value of x different from x=-1 will give us y value that is bigger then -4, therefore the parabola is located on and above x=-1. The equation of the line is y=-7, therefore the function will never intersect the parabola and there is no solution to the equations system.
If the parabola has a minimum point and b is above the vertex (b>y of the vertex) there are 2 intersection points between the parabola and the line, therefore there are 2 solutions to the systems of the equation.
Consider the following example:
How many intersection points has the system of equations y=(x+1)2-4 and y=7?
The vertex point of the parabola is (-1,-4), this is a minimum point since any value of x different from x=-1 will give us y value that is bigger then -4, therefore the parabola is located on and above x=-1. The equation of the line is y=7, therefore the function will intersect the parabola at 2 points and there are 2 solutions to the equations system.
The graph below shows the parabola and the different lines that were introduced in the example. The intersection points between the parabola and each one of the lines are marked on the graphs.
If the parabola has a maximum point and b is above the vertex (b>y of the vertex) there are no intersection points between the parabola and the line, therefore there are no solutions to the systems of the equation.
Consider the following example:
How many intersection points has the system of equations y=-(x-1)2+21 and y=30?
The vertex point of the parabola is (1,21), this is a maximum point since any value of x different from x=1 will give us y value that is smaller then 21, therefore the parabola is located on and below x=21. The equation of the line is y=30, therefore the function will never intersect the parabola and there is no solution to the equations system.
If the parabola has a maximum point and b is below the vertex (b<y of the vertex) there are 2 intersection points between the parabola and the line, therefore there are 2 solutions to the systems of the equation.
Consider the following example:
How many intersection points has the system of equations y=-(x-1)2+21 and y=10?
The vertex point of the parabola is (1,21), this is a maximum point since any value of x different from x=1 will give us y value that is smaller then 21, therefore the parabola is located on and below x=21. The equation of the line is y=10, therefore the function will intersect the parabola at 2 points and there are 2 solutions to the equations system.
The graphs below shows the parabola and the different lines that were introduced in the example. The intersection points between the parabola and each one of the lines are marked on the graphs.
Algebraical solution of linear and quadratic systems of equations
Algebraic solution of linear and quadratic systems of equations can be done by substitution method. In this method we substitute an expression containing x for y (plugging an expression instead of y), so we are left with one equation containing one variable x.
Solving a system of equations by substitution steps
Step 1: Isolate y in one of the equations so you have an equation containing y=expression with x. Note that it is the easiest to isolate y because we have x2 in one of the equations and isolating x will cause opening brackets of more complex equation like (y-a)2 in the next step.
Step 2: Substitute the expression from step 1 for a variable into the other equation. Now you stay with one equation that has only one variable that can be solved.
Step 3: Solve the equation for the remaining value and plug it into one of the equations to calculate the other variable.
For a detailed explanation about the ways of solving a quadratic equation learn the solving a quadratic equation topic.
Step 4: You can check your answer by plugging its x value into the equation you didn’t use in step 3 and making sure you are getting the same y value as in step 3.
Consider the following example:
Solve the system of equations y=3x2+5x+3 and x=y-10.
Step 1: Isolating y in the second equation:
x=y-10
y=x+10
Step 2: Plugging y=x+10 from the second equation as the substitute for y in the first equation getting
x+10=3x2+5x+3
Step 3: Solving the equation for x and then calculating y:
3x2+4x-7=0
x1=[-4+√(42-4*3*-7)]/(2*6)=(-4+√100)/6=(-4+10)/6=6/6=1
y1= x1+10=1+10=11 the first solution is (1,10)
x2=[-4–√(42-4*3*-7)]/(2*6)=(-4-√100)/6=(-4-10)/6=-14/6=-7/3=-21/3
y2= x2+10=-21/3 +10=72/3 the second solution is (-21/3, 72/3)
Step 4: Checking the solutions: To calculate the values of y we plugged the values of x1 and x2 into the second equation. Now we can plug x1 and x2 into the first equation and check if we get the same values of y:
y=3x2+5x+3
y1=3*12+5*1+3=3+5+3=11
y2=3*(-7/3)2+5*-7/3+3=(3*-7*-7)/(3*3)-35/3+3=+49/3-35-3+9/3=(+49-35+9)/3=23/3=7+2/3=72/3
Determining if a given solution is a solution for a system of equations
Each answer for a system of equations must satisfy both equations, therefore me must plug the given solution into each one of the equations. Plugging the solution into just one equation is not enough.
Graphically speaking, each point that is located on the graph of the first equation satisfies the first equation and each point that is located on the graph of the second equation satisfies the second equation, but only the points that are located on the intersection points of the two equations are the answers to the system of the equations.
Consider the following example:
The point (5,6):
Plugging the point (5,6) into the first equation:
y=x+1
6=5+1
6=6 is a true statement therefore we know that the solution (5,6) satisfies the first equation.
Plugging the point (5,6) into the second equation:
y=x2-2x+3
6=5*5-2*5+3
6=25-10+3
6=18 is a false statement therefore we know that the solution (5,6) doesn’t satisfy the second equation.
The answer is that the point (5,6) is not a solution for a system of the equations.
The point (1,2):
Plugging the point (1,2) into the first equation:
y=x+1
2=1+1
2=2 is a true statement therefore we know that the solution (1,2) satisfies the first equation.
Plugging the point (1,2) into the second equation:
y=x2-2x+3
2=1*1-2*1+3
2=1-2+3
2=2 is a true statement therefore we know that the solution (1,2) satisfies the second equation.
The answer is that the point (1,2) is a solution for a system of the equations.
You just finished studying linear and quadratic systems of equations topic, the third topic of passport to advanced mathematics subscore!
Continue studying the next passport to advanced mathematics subscore topic- exponential expressions.
