HA19 - Linear and quadratic simultaneous equations

HA19 - Linear and quadratic simultaneous equations

Can a line and a curve have more than one common point? HA19 uses substitution to find all coordinate pairs that satisfy a linear and a quadratic equation together, then explains what those pairs mean on a graph.

Substitution

Two simultaneous equations describe the same unknowns. A solution must make both equations true at once. When one relationship is quadratic, substitution can produce two possible values rather than the single value typical of two intersecting straight lines.

Substitution into a quadratic relationship

Substitution replaces a variable by an equal expression from the other equation. Use the linear equation to express one variable in terms of the other, substitute the complete expression, then solve the resulting quadratic. Recover the second variable for every real root.

If both equations give yy, their right-hand sides must be equal at a common point. If not, rearrange the linear equation first. Keep brackets around any substituted expression, especially if it is squared.

Worked example 1: both equations already give y

Question. Solve y=x+2y=x+2 and y=x2−4y=x^2-4 simultaneously.

Plan. Equate the two expressions for yy, solve for xx, then return to the linear rule.

x+2=x2−4⇒x2−x−6=0⇒(x−3)(x+2)=0.x+2=x^2-4 \quad\Rightarrow\quad x^2-x-6=0 \quad\Rightarrow\quad (x-3)(x+2)=0.

Hence x=3x=3 or x=−2x=-2. For each root,

xy=x+235−20\begin{array}{c|c} x&y=x+2\\\hline 3&5\\ -2&0 \end{array}

Answer. (x,y)=(3,5)(x,y)=(3,5) or (−2,0)(-2,0).

Check. In the quadratic equation, 32−4=53^2-4=5 and (−2)2−4=0(-2)^2-4=0, agreeing with the two linear outputs. Both complete pairs satisfy both original equations.

Now the linear equation needs rearranging before it can replace yy.

Worked example 2: rearrange before substituting

Question. Solve x+y=4x+y=4 and y=x2−2y=x^2-2.

Plan. Write y=4−xy=4-x, substitute into the quadratic relationship, and keep both roots.

4−x=x2−2⇒x2+x−6=0⇒(x+3)(x−2)=0.4-x=x^2-2 \quad\Rightarrow\quad x^2+x-6=0 \quad\Rightarrow\quad (x+3)(x-2)=0.

When x=2x=2, y=4−2=2y=4-2=2. When x=−3x=-3, y=4−(−3)=7y=4-(-3)=7.

Answer. (2,2)(2,2) and (−3,7)(-3,7).

Check. The sums are 2+2=42+2=4 and −3+7=4-3+7=4. Independently, the quadratic gives 22−2=22^2-2=2 and (−3)2−2=7(-3)^2-2=7.

The quadratic stage supplies the branches of the answer; each branch must travel back through an original relationship to become a complete pair.

Watch out

Solving the quadratic gives candidate values of one variable, not the complete answer. Do not stop before finding the matching value of the other variable for each root.

Use the same process with fresh coefficients.

Coordinate solutions

Writing two lists of values can lose which output belongs to which input. Use a row for each root, or write complete ordered pairs throughout.

Valid coordinate pairs

An ordered pair (x,y)(x,y) places the x-coordinate first. Every real root of the substituted quadratic must be considered, but each final pair must satisfy both original equations and any stated restrictions. Keep exact roots until after calculating their matching coordinates.

Factorisation may not work neatly; the quadratic formula still gives both roots. The signs in corresponding coordinates must stay linked.

Worked example 3: two exact surd pairs

Question. Solve x=2y−1x=2y-1 and x=y2−3x=y^2-3. Give exact coordinate pairs.

Plan. Equate the expressions for xx. Solve the quadratic in yy, then find xx from the linear equation.

2y−1=y2−3⇒y2−2y−2=0.2y-1=y^2-3 \quad\Rightarrow\quad y^2-2y-2=0.

The formula gives

y=2±(−2)2−4(1)(−2)2=2±122=1±3.y=\frac{2\pm\sqrt{(-2)^2-4(1)(-2)}}2 =\frac{2\pm\sqrt{12}}2=1\pm\sqrt3.

Using x=2y−1x=2y-1, the corresponding values are 1+231+2\sqrt3 and 1−231-2\sqrt3.

Answer. (1+23, 1+3)(1+2\sqrt3,\ 1+\sqrt3) and (1−23, 1−3)(1-2\sqrt3,\ 1-\sqrt3).

Check. For either consistently chosen sign,

(1±3)2−3=1±23,2(1±3)−1=1±23.(1\pm\sqrt3)^2-3=1\pm2\sqrt3, \qquad 2(1\pm\sqrt3)-1=1\pm2\sqrt3.

Both original expressions for xx agree. Mixing a plus choice in one coordinate with a minus choice in the other would fail the linear equation.

A context can rule out a mathematically correct pair. Apply that decision after finding all algebraic possibilities.

Worked example 4: an ordered length and width

Question. A rectangle has length xx cm and width yy cm, where x>y>0x>y>0. Its half-perimeter is nine centimetres and its area is fourteen square centimetres. Find its dimensions.

Plan. Solve x+y=9x+y=9, xy=14xy=14, then apply the stated length-width ordering.

From the linear equation, y=9−xy=9-x. Hence

x(9−x)=14⇒x2−9x+14=0⇒(x−2)(x−7)=0.x(9-x)=14 \quad\Rightarrow\quad x^2-9x+14=0 \quad\Rightarrow\quad (x-2)(x-7)=0.

The algebraic pairs are (2,7)(2,7) and (7,2)(7,2). The first violates x>yx>y; the second meets it and has two positive dimensions.

Answer. Length seven centimetres and width two centimetres.

Check. Half-perimeter 7+2=97+2=9, area 7×2=147\times2=14, and 7>2>07>2>0. The discarded pair solves the equations but not the complete stated problem.

The equations alone describe the algebraic possibilities. An extra statement in the problem can select among those possibilities, so keep that decision visible.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account