Solving systems of linear and quadratic equations worksheet answers

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You have probably solved systems of linear equations. But what about a system of two equations where one equation is linear, and the other is quadratic?

We can use a version of the substitution method to solve systems of this type.

Remember that the slope-intercept form of the equation for a line is y=mx+b, and the standard form of the equation for a parabola with a vertical axis of symmetry is y=ax2+bx+c,a≠0.

To avoid confusion with the variables, let us write the linear equation as y=mx +d where m is the slope and d is the y-intercept of the line.

Substitute the expression for y from the linear equation, in the quadratic equation. That is, substitute mx+d  for y in y=ax2+bx+c .

mx+d=ax2+bx+c

Now, rewrite the new quadratic equation in standard form.

Subtract mx+d  from both sides.

(mx+d)− (mx+d)=(ax2+bx+c)−(mx+d)0=ax2+(b−m)x+( c−d)

Now we have a quadratic equation in one variable, the solution of which can be found using the quadratic formula.

The solutions to the equation ax2+(b−m)x+(c− d)=0   will give the x-coordinates of the points of intersection of the graphs of the line and the parabola. The corresponding y-coordinates can be found using the linear equation.

Another way of solving the system is to graph the two functions on the same coordinate plane and identify the points of intersection.

Example 1:

Find the points of intersection between the line y=2x+1  and the parabola y=x2−2.

Substitute 2x+1  for y in y=x2−2.

2x+1= x2−2

Write the quadratic equation in standard form.

2x+1−2x−1=x2−2−2x−10=x2−2x−3

Use the quadratic formula to find the roots of the quadratic equation.

Here, a=1,b=−2, and c=−3.

x=−(−2)± (−2)2−4(1)(−3)2(1)=2±4+12 2=2±42=3,−1

Substitute the x-values in the linear equation to find the corresponding y-values.

x=3 ⇒y=2(3)+1=7 x=−1⇒y=2(−1)+1 =−1
Therefore, the points of intersection are (3,7)  and (−1,−1).

Graph the parabola and the straight line on a coordinate plane.

A similar method can be used to find the intersection points of a line and a circle.

Example 2:

Find the points of intersection between the line y=−3x  and the circle x2 +y2=3.

Substitute −3x  for y in x2+y2=3 .

x2+(−3x)2=3

Simplify.

x2+9x2=310x2=3x2=310
Taking square roots, x=±3 10.

Substitute the x-values in the linear equation to find the corresponding y-values.
x=310⇒y=−3(310) = −3310x=−310⇒y=−3(−310) =3 310

Therefore, the points of intersection are (310,−3310)  and (− 310,3310).

Graph the circle and the straight line on a coordinate plane.

...or a line and an ellipse.

Example 3:

Solve the system of equations y=−5  and x29+y24 =1.

Substitute −5  for y in −5.

x29+(−5)24=1

Simplify.

x2 9+(−5)24=14x236+9(25)36=14x2+225 =364x2=−189x2=−1894

Here we have a negative number as the square of a number. So, the two equations do not have real solutions.

Graph the ellipse and the straight line on a coordinate plane.

We can see that the two do not intersect.

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Linear and quadratic equations are the algebraic systems of equations consisting of one linear equation and one quadratic equation. The goal of solving systems linear quadratic equations is to significantly reduce two equations having two variables down to a single equation with only one variable. Since each equation in the system consists of two variables, one way to decrease the number of variables in an equation is by substituting an expression for a variable.

In a test, you will be expected to identify the solution(s) to systems either algebraically or graphically.

Examples of Linear Equation and Quadratic Equation

Following are the example of Linear and quadratic equations:

y = x + 1

y = x² −1

In systems linear quadratic equations, both equations are simultaneously true. Simply to say, because the 1st equation tells us that y is equivalent to x + 1, the y in the 2nd equation is also equivalent to x + 1. Hence, we can plug in x + 1 as a substitute for y in the 2nd equation:

y = x² − 1

x + 1 = x² −1

From here, we are able to solve and simplify the quadratic equation for x, which provides us with the x-values of the solutions to the linear-quadratic system. Then, we can also use the x-values and either equation in the system in order to find out the y-values.

Solving Linear-Quadratic Systems

You might have solved systems of linear equations. But what about a system of 2 equations where 1 equation is linear and the remaining is quadratic?

We can apply a version of the substitution technique in order to solve systems of this type.

Note that the standard form of the equation for a parabola with a vertical axis of symmetry is y = ax² + bx + c, a ≠ 0 and the slope-intercept form of the equation for a line is y = mx + b,

To avoid any sort of confusion with the variables, write the linear equation as y= mx+d where,

m = The slope.

d = The y-intercept of the line.

Substitute the expression for y from the linear equation, into the quadratic equation. In other words, substitute mx + d for y in y = ax² + bx + c.

mx + d = ax² + bx + c

Now, rewrite the new quadratic equation in the standard form.

Subtract mx+d from both sides of the equation

(mx + d) − (mx + d) = (ax² + bx + c) − (mx + d)0 = ax² + (b − m)x + (c − d)

Now we have a quadratic equation in 1 variable, the solution of which can be determined with the help of the quadratic formula.

The solutions to the equation ax² + (b − m)x + (c − d) = 0 will provide us with the x-coordinates of the points of intersection of the graphs of the parabola and the line. The corresponding y-coordinates can be identified with the help of the linear equation.

Another way of solving the system is to graph the two functions on a similar coordinate plane and then determine the points of intersection.

Solved Examples

Example:

Determine the points of bisection between the line y = 2x + 1 and the parabola y = x² − 2.

Solution:

Substitute the values in the equation 2x + 1 for y in y = x² − 2.

2x + 1 = x² − 2

Now, express the quadratic equation in standard form.

2x + 1− 2x − 1 = x² − 2 − 2x − 10 = x² − 2x − 3

Apply the quadratic formula in order to identify the roots of the quadratic equation.

Here,

a = 1

b = −2

c = −3

x = √[− (−2) ± (−2)2 − 4(1)(−3)]/2(1)

= √[2 ± 4 + 12]/2

= 2 ± 4/2

= 3, −1

Then, Substitute the x-values in the linear equation in order to identify the corresponding y-values.

x = 3 ⇒ y = 2(3) + 1 = 7

x = −1 ⇒ y = 2(−1) + 1 = −1

Thus, the points of bisection are (3, 7) and (−1, −1).

Also, remember that the same method can be used to determine the intersection points of a line and a circle.

Check below the Graphing of the parabola and the straight line on a coordinate plane.

Example: Determine the points of intersection between the line y = −3x and the circle x² + y² = 3.

Solution:

Substitute −3x for y in x² + y² = 3.

x² + (−3x)² = 3

Simplify the system of quadratic equations

x² + 9x² = 3

10x² = 3

x² = 3/10

Taking square roots, x = ±√(3/10).

Now, Substituting the x-values in the linear equation in order to identify the corresponding y-values.

x = √(3/10) ⇒ y = −3(√(3/10))

=−(3/3√10)

x = −√(3/10) ⇒ y =−3(−√(3/10))

= 3/3√10

Hence, the points of intersection come out to be (√(3/10), (-3√3/10), and (−3/√10√, 3√3/√10).

Refer below for the Graphing of the circle and the straight line on a coordinate plane.

Fun Facts

• While solving quadratic simultaneous equations, if we get a negative number as the square of a number in the outcome, then the two equations do not have real solutions.

• A  quadratic and linear system can be represented by a line and a parabola in the xy-plane.

• Each intersection of the line and the parabola depicts a solution to a linear-quadratic system.

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