Systems of Equations, Worked Examples (KBAT)

Make $x$ the subject of the linear equation:

$$x=y+1$$

Substitute into $x^{2}+xy=6$. Expand both $(y+1)^{2}$ and the product $(y+1)y$:

$$(y+1)^{2}+(y+1)y=6 \;\Rightarrow\; y^{2}+2y+1+y^{2}+y=6$$

Collect like terms and bring everything to one side:

$$2y^{2}+3y+1=6 \;\Rightarrow\; 2y^{2}+3y-5=0$$

Factorise the quadratic. The leading coefficient is $2$, so split the middle term: $2y^{2}+5y-2y-5$, which factors as

$$(2y+5)(y-1)=0 \;\Rightarrow\; y=-\tfrac{5}{2} \ \text{or}\ y=1$$

Substitute each root back into $x=y+1$:

$$y=1 \Rightarrow x=2, \qquad y=-\tfrac{5}{2} \Rightarrow x=-\tfrac{3}{2}$$

Label the equations: $(1)\;x+y+z=4$, $(2)\;2x+3y-z=-2$, $(3)\;3x-y+2z=13$. Eliminate $z$ first.

Add $(1)$ and $(2)$, whose $z$-terms are $+z$ and $-z$:

$$(1)+(2):\; 3x+4y=2 \quad (4)$$

Now eliminate $z$ again from a different pair. Multiply $(2)$ by $2$ to match the $+2z$ in $(3)$, then add:

$$2\times(2):\; 4x+6y-2z=-4$$

$$[2\times(2)]+(3):\; 7x+5y=9 \quad (5)$$

Solve $(4)$ and $(5)$ by elimination. Multiply $(4)$ by $5$ and $(5)$ by $4$ so the $y$-terms match:

$$15x+20y=10, \qquad 28x+20y=36$$

Subtract the first from the second:

$$13x=26 \;\Rightarrow\; x=2$$

Substitute $x=2$ into $(4)$, then find $z$ from $(1)$:

$$3(2)+4y=2 \Rightarrow y=-1, \qquad z=4-x-y=4-2-(-1)=3$$

A tangent meets the curve at exactly one point, so the equation formed by solving the line and curve together must have one repeated root. Set the two expressions for $y$ equal:

$$x^{2}+3=2x+c$$

Rearrange into a quadratic in $x$ with everything on one side:

$$x^{2}-2x+(3-c)=0$$

For one repeated root the discriminant $b^{2}-4ac$ must be zero. Here $a=1$, $b=-2$, $c_{\text{quad}}=3-c$:

$$(-2)^{2}-4(1)(3-c)=0 \;\Rightarrow\; 4-12+4c=0$$

Solve for $c$:

$$4c=8 \;\Rightarrow\; c=2$$

Put $c=2$ back into the quadratic to find the point of contact. It becomes a perfect square:

$$x^{2}-2x+1=0 \;\Rightarrow\; (x-1)^{2}=0 \;\Rightarrow\; x=1$$

Find $y$ from the line $y=2x+c=2x+2$:

$$y=2(1)+2=4$$

Let the length be $x$ m and the width be $y$ m. Halving the perimeter gives the sum of the two sides, and the area gives their product:

$$x+y=11, \qquad xy=24$$

A pair of numbers with a known sum and a known product are the two roots of the quadratic $t^{2}-(\text{sum})t+(\text{product})=0$. Substitute the values in:

$$t^{2}-11t+24=0$$

Factorise:

$$(t-3)(t-8)=0 \;\Rightarrow\; t=3 \ \text{or}\ t=8$$

The two roots $3$ and $8$ are the width and length in some order. Since the length is greater than the width:

$$\text{length}=8, \qquad \text{width}=3$$

Let the two numbers be $x$ and $y$. The first condition gives the sum directly, and the second gives an equation in $\frac{1}{x}$ and $\frac{1}{y}$:

$$x+y=5, \qquad \frac{1}{x}+\frac{1}{y}=\frac{5}{6}$$

Combine the fractions on the left of the second equation over a common denominator $xy$:

$$\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}=\frac{x+y}{xy}$$

Substitute $x+y=5$ into this combined fraction and equate it to $\frac{5}{6}$:

$$\frac{5}{xy}=\frac{5}{6} \;\Rightarrow\; xy=6$$

Now $x$ and $y$ are the two roots of the quadratic with sum $5$ and product $6$:

$$t^{2}-5t+6=0 \;\Rightarrow\; (t-2)(t-3)=0 \;\Rightarrow\; t=2 \ \text{or}\ t=3$$

The perimeter gives a linear equation in $x$ and $y$, and Pythagoras' theorem on the right angle gives a second, non-linear equation:

$$x+y+10=24 \;\Rightarrow\; x+y=14, \qquad x^{2}+y^{2}=10^{2}=100$$

Substituting $y=14-x$ directly would work, but it is faster to use the identity $(x+y)^{2}=x^{2}+2xy+y^{2}$ to find the product $xy$ first:

$$14^{2}=100+2xy \;\Rightarrow\; 196-100=2xy$$

Solve for $xy$:

$$2xy=96 \;\Rightarrow\; xy=48$$

$x$ and $y$ are now the two roots of the quadratic with sum $14$ and product $48$:

$$t^{2}-14t+48=0 \;\Rightarrow\; (t-6)(t-8)=0 \;\Rightarrow\; t=6 \ \text{or}\ t=8$$

Since $x>y$, assign the larger root to $x$:

$$x=8, \qquad y=6$$

Let the price of one apple, orange and pear be RM$a$, RM$o$ and RM$p$ respectively. The three purchases give three equations; label them:

$$(1)\; 2a+o+3p=22, \quad (2)\; a+3o+2p=21, \quad (3)\; 3a+2o+p=17$$

$p$ has coefficient $1$ in equation $(3)$, so make it the subject and eliminate it from $(1)$ and $(2)$:

$$p=17-3a-2o$$

Substitute into $(1)$:

$$2a+o+3(17-3a-2o)=22 \;\Rightarrow\; 7a+5o=29 \quad (4)$$

Substitute into $(2)$:

$$a+3o+2(17-3a-2o)=21 \;\Rightarrow\; 5a+o=13 \quad (5)$$

Make $o$ the subject of $(5)$ and substitute into $(4)$:

$$o=13-5a, \qquad 7a+5(13-5a)=29 \;\Rightarrow\; -18a=-36 \;\Rightarrow\; a=2$$

Back-substitute to find $o$ and $p$:

$$o=13-5(2)=3, \qquad p=17-3(2)-2(3)=5$$

Two distinct points of intersection mean the equation formed by solving the line and curve together has two distinct real roots, so its discriminant must be positive rather than zero. Set the two expressions for $y$ equal:

$$x+k=x^{2}-3x+7$$

Rearrange into a quadratic in $x$ with everything on one side:

$$x^{2}-4x+(7-k)=0$$

For two distinct real roots, $b^{2}-4ac>0$. Here $a=1$, $b=-4$, $c=7-k$:

$$(-4)^{2}-4(1)(7-k)>0$$

Simplify the inequality:

$$16-28+4k>0 \;\Rightarrow\; 4k>12 \;\Rightarrow\; k>3$$