Question

Use the X and Y intercept to work out these simultaneous equations.

1. y = x + 5 and y = 2x + 1

2. y = x + 4 and y = 3x - 2

3. y = x - 3 and y = 2x - 7

4. y = 3x - 7 and y = x-1​

Answer 1

Answer:

option 4•)y=3x-7 and y=x-1

Answer 2

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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                       =7x=−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=±310−−√.

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

x=310−−√⇒y=−3(310−−√)                              =−33√10√x=−310−−√⇒y=−3(−310−−√)                                  =33√10

Therefore, the points of intersection are (3√10√,  −33√10) and (−3√10√,  33√10).

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.

x29+(−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.