28. Simultaneous Equations (Linear and Non-Linear)

Question 1

Solve the simultaneous equations:

y=-x+2\\

y=x^2 + 2x - 1

ANSWER: Multiple Answers (Type 1)

Answer:

$x_1=-3.79$ , $y_1=5.79$

$x_2=0.791,$ , $y_2=1.21$

Workings:

$-x+2=x^2+2x-1$

$x^2+3x-3=0$

Putting into quadratic formula:

$x= \dfrac{-3\pm \sqrt{21}}{2}$

This gives:

$x_1=-3.79$ OR $x_2=0.791$

Substituting these values into original equations gives:

$y_1=5.79$ OR $y_2=1.21$

$(-3.79, 5.79)$ OR $(0.791, 1.21)$

Marks = 4

Question 2

Solve the simultaneous equations:

x^2+y^2=9\\

2y=x+1

ANSWER: Multiple Answers (Type 1)

Answer:

$x_1=-2.85$ , $y_1=-0.927$

$x_2=2.45$ , $y_2=1.73$

Workings:

Substituting equation two into equation one:

$x^2+(\dfrac{x+1}{2})^2-9=0$

$x^2+\dfrac{x^2+2x+1}{4}-9=0$

$4x^2+x^2+2x+1-36=0$

$5x^2+2x-35=0$

Putting into quadratic formula:

$x = \dfrac{-2\pm \sqrt{704}}{10}$

This gives:

$x_1 = -2.85$ OR $x_2 = 2.45$

Substituting these values into original equations gives:

$y_1=-0.927$ OR $y_2 = 1.73$

$(-2.85,-0.927)$ OR $(2.45, 1.73)$

Marks = 4

ALT SOLUTION:

Rearranging equation two gives $x=2y-1$

Substituting this into equation one gives:

$(2y-1)^2+y^2=9$

$4y^2-2y-2y+1+y^2=9$

$5y^2-4y-8=0$

Using the quadratic formula:

$y=\dfrac{-(-4) \pm \sqrt{(-4)^2 - (4 \times 5 \times (-8))}}{2 \times 5}$

$y=\dfrac{4 \pm \sqrt{176}}{10}$

$y_1 = 1.73, \quad y_2 = -0.927$

Substituting these values of $y$ back into equation two gives:

$x_1 = 2.46, \quad x_2 = -2.85$

Question 3

Solve the simultaneous equations:

3x+y=-9\\

x^2+2x-3 =y

ANSWER: Multiple Answers (Type 1)

Answer:

$x_1=-3$ , $y_1=0$

$x_2=-2$ , $y_2=-3$

Workings:

Substitute equation two into equation one:

$3x+x^2+2x-3=-9$

$x^2+5x+6=0$

$(x+3)(x+2)=0$

This gives:

$x_1=-3$ OR $x_2=-2$

Substituting these values into the original equation gives:

$y_1=0$ OR $y_2=-3$

$(-3, 0)$ OR $(-2, -3)$

Marks = 4

Question 4

Solve the simultaneous equations:

y=3x -1\\

3x^2+2y^2=35

ANSWER: Multiple Choice (Type 2)

A: $x_1=-1$ , $y_1=-4$ , $x_2=\dfrac{11}{7}$ , $y_2=\dfrac{26}{7}$

B: $x_1=-1$ , $y_1=-6$ , $x_2=\dfrac{15}{7}$ , $y_2=\dfrac{23}{7}$

C: $x_1=-1$ , $y_1=-4$ , $x_2=\dfrac{15}{7}$ , $y_2=\dfrac{26}{7}$

D: $x_1=-1$ , $y_1=-6$ , $x_2=\dfrac{11}{7}$ , $y_2=\dfrac{23}{7}$

Answer: A

Workings:

Substitute equation one into equation two:

$3x^2+2(3x-1)^2-35=0$

$21x^2-12x-33=0$

$7x^2-4x-11=0$

$(7x-11)(x+1)=0$

This gives:

$x_1=\dfrac{11}{7}$ OR $x_2=-1$

Substituting these values into the original equations gives:

$y_1=\dfrac{26}{7}$ OR $y_2=-4$

$(\dfrac{11}{7}, \dfrac{26}{7})$ OR $(-1, -4)$

Marks = 5

Question 5

By eliminating $y$ from the following equations

y=2-4x\\

3x^2+xy+11=0

show that $x^2-2x-11=0$ .

Hence or otherwise, solve the simultaneous equations, giving your answers

in the form $a+b\sqrt3$ , where $a$ and $b$ are integers.

ANSWER: Multiple Choice (Type 1)

A: $x_1=7+4\sqrt{3}$ , $y_1 = -3-3\sqrt{3}$ OR $x_2=5-3\sqrt{3}$ , $y_2=-6+3\sqrt{3}$

B: $x_1=1+2\sqrt{3}$ , $y_1 = -2-8\sqrt{3}$ OR $x_2=1-2\sqrt{3}$ , $y_2=-2+8\sqrt{3}$

C: $x_1=7+9\sqrt{3}$ , $y_1 = -7-5\sqrt{3}$ OR $x_2=9-5\sqrt{3}$ , $y_2=-5+7\sqrt{3}$

D: $x_1=8+2\sqrt{3}$ , $y_1 = -7-3\sqrt{3}$ OR $x_2=4-4\sqrt{3}$ , $y_2=-4+2\sqrt{3}$

Answer: B

Workings:

Substitute equation one into equation two:

$3x^2+x(2-4x)+11=0$

$-x^2+2x+11=0$

$x^2-2x-11=0$

Substitute into quadratic formula:

$x=\dfrac{2\pm \sqrt{48}}{2}$

$x=\dfrac{2\pm 4\sqrt{3}}{2}$

$x={1\pm 2\sqrt{3}}$

This gives:

$x_1=1+2\sqrt{3}$ OR $x_2=1-2\sqrt{3}$

Substituting these values into the original equations gives:

$y_1=-2-8\sqrt{3}$ OR $y_2=-2+8\sqrt{3}$

$(1+2\sqrt{3}, -2-8\sqrt{3})$ OR $(1-2\sqrt{3}, -2+ 8\sqrt{3})$

Marks = 7

Question 6 REMAKE

Given the following simultaneous equations

x-y=3\sqrt{2}

x^2+y^2-9=0

Find the values of $x$ and $y$

ANSWER: Multiple Choice (Type 1)

A: $x=\dfrac{3\sqrt{2}}{2}, \quad y=-\dfrac{3\sqrt{2}}{2}$

B: $x=\dfrac{2\sqrt{2}}{3}, \quad y=-\dfrac{2\sqrt{2}}{3}$

C: $x=\dfrac{3\sqrt{2}}{4}, \quad y=-\dfrac{3\sqrt{2}}{4}$

D: $x=\dfrac{4\sqrt{2}}{3}, \quad y=-\dfrac{4\sqrt{2}}{3}$

Answer: D

Workings:

$x-y=3\sqrt{2}$

So $y=x-3\sqrt{2}$

Substituting the expression for $y$ into the second equation:

\begin{aligned}x^2+y^2-9 &=0 \\[1.5em] x^2 +(x-3\sqrt{2})(x-3\sqrt{2})-9 &= 0 \\[1.5em] x^2 + x^2-3\sqrt{2}x-3\sqrt{2}x+18-9 &= 0 \\[1.5em] 2x^2-6\sqrt{2}x+9 &=0\end{aligned}

Using the quadratic formula the find the value of $x$ :

\begin{aligned}x &=\dfrac{-(-6\sqrt{2}) \pm \sqrt{(-6\sqrt{2})^2 - (4 \times 2 \times 9)}}{2 \times 2} \\[1.5em] &=\dfrac{6\sqrt{2} \pm \sqrt{72 - 72}}{4} \\[1.5em] &= \dfrac{6\sqrt{2}}{4} \\[1.5em] &= \dfrac{3\sqrt{2}}{2}\end{aligned}

Finally, substituting the value of $x$ into the original equation gives the value of $y$ :

\begin{aligned}y &= x-3\sqrt{2} \\[1.5em] y &= \dfrac{3\sqrt{2}}{2}-3\sqrt{2} \\[1.5em] y& =-\dfrac{3\sqrt{2}}{2}\end{aligned}

Marks = 6