Linear-Quadratic Systems Practice Problems with Solutions

Master solving systems where one equation is linear and one is quadratic using substitution method. Practice with step-by-step solutions and examples.

📚What You'll Master in This Practice Session
  • Solve linear-quadratic systems using the substitution method effectively
  • Find intersection points between parabolas and straight lines graphically
  • Apply the quadratic formula to solve resulting single-variable equations
  • Identify systems with zero, one, or two solution points
  • Substitute solutions back to find complete coordinate pairs
  • Interpret solutions as geometric intersections of curves and lines

Understanding Linear-Quadraric Systems of Equations

Complete explanation with examples

Solving a system of equations when one of them is linear and the other is quadratic

When we have a system of equations where one of the equations is linear and the other quadratic
we will use the substitution method:

We will isolate one variable from an equation, place in the second equation the value of the expression of the variable we have isolated, and in this way, we will obtain an equation with one variable. We will solve for XX or YY and then place it in one of the original equations to find the complete point. The point we discover will be the point of intersection of the line with the parabola, and it will also be the solution of the system of equations.

Graphical comparison of linear-quadratic systems showing three cases: two points of intersection (two solutions), one point (one solution), and no intersection (no solution).

Detailed explanation

Practice Linear-Quadraric Systems of Equations

Test your knowledge with 3 quizzes

Which formula represents line 1 in the graph below?

BBBCCC12

Examples with solutions for Linear-Quadraric Systems of Equations

Step-by-step solutions included
Exercise #1

The following functions are graphed below:

f(x)=x26x+8 f(x)=x^2-6x+8

g(x)=4x17 g(x)=4x-17

For which values of x is
f(x)<0 true?

BBBAAAKKK

Step-by-Step Solution

To solve the inequality f(x)=x26x+8<0 f(x) = x^2 - 6x + 8 < 0 , we follow these steps:

  • Step 1: Find the roots of the equation f(x)=0 f(x) = 0 using the factoring method.
  • Step 2: Determine the intervals formed by these roots.
  • Step 3: Test points from each interval to determine where f(x)<0 f(x) < 0 .

Step 1: Factor the quadratic equation x26x+8=0 x^2 - 6x + 8 = 0 .
Factoring gives: (x2)(x4)=0(x - 2)(x - 4) = 0.

Thus, the roots are x=2 x = 2 and x=4 x = 4 .

Step 2: The roots divide the number line into three intervals: x<2 x < 2 , 2<x<4 2 < x < 4 , and x>4 x > 4 .

Step 3: Choose a test point from each interval and plug it into f(x) f(x) :

  • For x<2 x < 2 , choose x=1 x = 1 : f(1)=126(1)+8=16+8=3 f(1) = 1^2 - 6(1) + 8 = 1 - 6 + 8 = 3 , which is positive, so f(x)0 f(x) \geq 0 .
  • For 2<x<4 2 < x < 4 , choose x=3 x = 3 : f(3)=326(3)+8=918+8=1 f(3) = 3^2 - 6(3) + 8 = 9 - 18 + 8 = -1 , which is negative, so f(x)<0 f(x) < 0 .
  • For x>4 x > 4 , choose x=5 x = 5 : f(5)=526(5)+8=2530+8=3 f(5) = 5^2 - 6(5) + 8 = 25 - 30 + 8 = 3 , which is positive, so f(x)0 f(x) \geq 0 .

Therefore, the interval where f(x)<0 f(x) < 0 is 2<x<4 2 < x < 4 .

The correct choice is:

2<x<4 2 < x < 4

Answer:

2 < x < 4

Video Solution
Exercise #2

Look at the graph below of the following functions:

f(x)=x26x+8 f(x)=x^2-6x+8

g(x)=x+4 g(x)=-x+4

For which values of x is
g(x)>0 true?

BBBCCC

Step-by-Step Solution

To solve the problem of finding for which values of x x , the function g(x)=x+4 g(x) = -x + 4 is greater than zero, we begin as follows:

  • Step 1: Set up the inequality g(x)>0 g(x) > 0 . This translates to x+4>0 -x + 4 > 0 .
  • Step 2: Solve the inequality:
    • Subtract 4 from both sides: x>4-x > -4.
    • Multiply both sides by 1-1, reversing the inequality: x<4x < 4.

Therefore, the solution to the problem is that g(x)>0 g(x) > 0 when x<4 x < 4 .

The corresponding choice that reflects this solution is choice 4: x<4 x < 4 .

Answer:

x<4

Video Solution
Exercise #3

The following function is graphed below:

f(x)=x26x+8 f(x)=x^2-6x+8

g(x)=x+4 g(x)=-x+4

For which values of x is
f(x) > g(x) true?

BBBCCC

Step-by-Step Solution

To determine for which values of x x the condition f(x)>g(x) f(x) > g(x) holds, follow these steps:

  • Step 1: Set up the inequality x26x+8>x+4 x^2 - 6x + 8 > -x + 4 .
  • Step 2: Rearrange all terms to one side: x26x+8+x4>0 x^2 - 6x + 8 + x - 4 > 0 simplifies to x25x+4>0 x^2 - 5x + 4 > 0 .
  • Step 3: Solve the related quadratic equation x25x+4=0 x^2 - 5x + 4 = 0 . This factors as (x1)(x4)=0 (x - 1)(x - 4) = 0 , giving roots x=1 x = 1 and x=4 x = 4 .
  • Step 4: Determine intervals defined by the roots: (,1) (-\infty, 1) , (1,4) (1, 4) , and (4,) (4, \infty) .
  • Step 5: Test points from each interval in the inequality x25x+4>0 x^2 - 5x + 4 > 0 :
    • For x=0 x = 0 (from interval (,1) (-\infty, 1) ), x25x+4=4 x^2 - 5x + 4 = 4 which is greater than 0.
    • For x=2 x = 2 (from interval (1,4) (1, 4) ), x25x+4=2 x^2 - 5x + 4 = -2 which is less than 0.
    • For x=5 x = 5 (from interval (4,) (4, \infty) ), x25x+4=9 x^2 - 5x + 4 = 9 which is greater than 0.
  • Step 6: From the test points, determine f(x)>g(x) f(x) > g(x) in intervals (,1) (-\infty, 1) and (4,) (4, \infty) .

Thus, f(x)>g(x) f(x) > g(x) for x<1 x < 1 or x>4 x > 4 .

Therefore, the solution to the problem is x<1,4<x x < 1, 4 < x , corresponding to choice 2.

Answer:

x < 1,4 < x

Video Solution
Exercise #4

Choose the formula that describes graph 1:

BBBAAAKKK12

Step-by-Step Solution

To solve this problem, we need to determine whether the provided graph corresponds to a quadratic or linear function.

  • First, we observe the shape of the graph.
  • The graph shows a downward curve, indicating it is a parabola.
  • The general form of a quadratic equation (y=ax2+bx+c)(y = ax^2 + bx + c) implies graph types.
  • Let's verify if one of the given quadratic choices represents this graph.

Since the graph is a parabola opening upwards, we'll evaluate the given quadratic equation y=x26x+8 y = x^2 - 6x + 8 . Analyzing it and comparing leads to:

  • The vertex form can be rewritten or identified mathematically or visually from the given expression.
  • This quadratic formula aligns with prominent features of the parabola: its vertex, orientation, and intercepts, matching the graph.

Thus, as the parabola aligns perfectly with quadratic properties such as opening upwards, the formula that describes graph 1 correctly is:
y=x26x+8 y = x^2 - 6x + 8 .

Answer:

y=x26x+8 y=x^2-6x+8

Video Solution
Exercise #5

The following functions are graphed below:

f(x)=x26x+8 f(x)=x^2-6x+8

g(x)=4x17 g(x)=4x-17

For which values of x is

f(x)>0 true?

BBBAAAKKK

Step-by-Step Solution

To solve the inequality f(x) > 0 , we first need to find the roots of the equation f(x)=x26x+8 f(x) = x^2 - 6x + 8 .

1. Find the roots of the quadratic equation:
The quadratic is x26x+8 x^2 - 6x + 8 . This can be factored into:
f(x)=(x2)(x4)=0 f(x) = (x - 2)(x - 4) = 0 .

2. Calculate the roots:
Setting each factor equal to zero gives the roots x=2 x = 2 and x=4 x = 4 .

3. Determine the intervals defined by these roots:
The roots divide the x-axis into three intervals: (,2) (-\infty, 2) , (2,4) (2, 4) , and (4,) (4, \infty) .

4. Test points in each interval to decide positivity:
- For x < 2 , select x=1 x = 1 : f(1) = 1^2 - 6(1) + 8 = 3 > 0 . Thus, f(x) > 0 in (,2) (-\infty, 2) .
- For 2 < x < 4 , select x=3 x = 3 : f(3) = 3^2 - 6(3) + 8 = -1 < 0 . Thus, f(x) < 0 in (2,4) (2, 4) .
- For x > 4 , select x=5 x = 5 : f(5) = 5^2 - 6(5) + 8 = 3 > 0 . Thus, f(x) > 0 in (4,) (4, \infty) .

Therefore, the solution to f(x) > 0 is when x < 2 or x > 4 .

The final solution is: x < 2, 4 < x .

Answer:

x < 2, 4 < x

Video Solution

Frequently Asked Questions

What is a linear-quadratic system of equations?

+
A linear-quadratic system consists of one linear equation (like y = 2x + 3) and one quadratic equation (like y = x² + 2x + 7). The solution represents where a straight line intersects with a parabola on a coordinate plane.

How do you solve linear-quadratic systems step by step?

+
Follow these steps: 1) Isolate one variable from the linear equation, 2) Substitute this expression into the quadratic equation, 3) Solve the resulting quadratic equation using factoring or the quadratic formula, 4) Substitute each x-value back to find corresponding y-values.

Why do linear-quadratic systems have 0, 1, or 2 solutions?

+
The number of solutions depends on how many times the line intersects the parabola. A line can miss the parabola entirely (0 solutions), touch it at exactly one point as a tangent (1 solution), or cross through it at two distinct points (2 solutions).

When should I use substitution method for linear-quadratic systems?

+
Always use substitution method for linear-quadratic systems. It's the most efficient approach because you can easily isolate a variable from the linear equation and substitute it into the quadratic equation, reducing the system to a single quadratic equation.

What does it mean when a linear-quadratic system has no solution?

+
No solution means the line and parabola don't intersect anywhere on the coordinate plane. Algebraically, this occurs when the substituted quadratic equation has no real solutions (negative discriminant in the quadratic formula).

How do I check if my linear-quadratic system solutions are correct?

+
Substitute both x and y coordinates of each solution point back into both original equations. If both equations are satisfied by the coordinates, your solutions are correct. Each solution should work in both the linear and quadratic equations.

What's the difference between solving linear systems and linear-quadratic systems?

+
Linear systems typically have one unique solution (intersection of two lines), while linear-quadratic systems can have 0, 1, or 2 solutions since a line can intersect a parabola at multiple points. Linear-quadratic systems also require solving a quadratic equation as an intermediate step.

Can elimination method be used for linear-quadratic systems?

+
While technically possible, elimination method is not recommended for linear-quadratic systems. Substitution method is much more straightforward because the linear equation easily provides an expression for one variable that can be directly substituted into the quadratic equation.

Continue Your Math Journey

Practice by Question Type