H2 Math Systems of Linear Equations: Study Guide & Examples

JC1 · H2 Mathematics · Study guide

Form and solve systems from purchases, stationary points and recurrence data; handle substitutions and exceptional parameter cases completely.

Concepts · worked methods · graphs7 min read · practice linked below

Define the unknowns before forming equations

Use one unknown for each quantity, with units. Translate each independent piece of information into an equation. Keep the unknowns in the same order in every row before entering coefficients in the simultaneous equation solver.

Prices, comparisons and discounts

Three pieces of information give three equations

A shop sells three blends of tea at $a,b,c$ dollars per kilogram. A customer pays $116 for 2 kg of A, 1 kg of B and 3 kg of C. One kilogram each of A and C costs $12 more than 1 kg of B. After discounts of 20%, 10% and 5% respectively, buying 1 kg of each costs $57.40. Find the original prices.

“$12 more” means $a+c=b+12$ . A 20% discount leaves 80% of the price, so

\begin{aligned}2a+b+3c&=116,\quad\text{(1)}\\a-b+c&=12,\quad\text{(2)}\\0.8a+0.9b+0.95c&=57.4.\quad\text{(3)}\end{aligned}

Eliminate $b$ by using $b=a+c-12$ . Substitute in (1), and multiply (3) by 20 first to avoid decimals:

\begin{aligned}3a+4c&=128,\quad\text{(4)}\\16a+18b+19c&=1148,\\34a+37c&=1364.\quad\text{(5)}\end{aligned}

Eliminate $a$ using $34(4)-3(5)$ :

\begin{aligned}25c&=260,\\c&=10.4,\\3a&=128-4(10.4),\\a&=28.8,\\b&=28.8+10.4-12\\&=27.2.\end{aligned}

The prices are $28.80, $27.20 and $10.40 per kilogram.

\begin{aligned}2(28.8)+27.2+3(10.4)&=116,\\0.8(28.8)+0.9(27.2)\\\quad+0.95(10.4)&=57.4.\end{aligned}

The simultaneous equation solver gives the same results from coefficient rows $(2,1,3;116)$ , $(1,-1,1;12)$ and $(0.8,0.9,0.95;57.4)$ . Display the equations even when the calculator performs the solution.

A stationary point supplies two conditions

Find a quadratic from a point and its minimum

The curve $y=ax^2+bx+c$ passes through $(-1,18)$ and has a stationary point at $(2,0)$ . Find its equation.

The stationary point lies on the curve, and its gradient is zero. These are separate conditions.

\begin{aligned}a-b+c&=18,\quad\text{(1)}\\4a+2b+c&=0,\quad\text{(2)}\\\frac{dy}{dx}&=2ax+b,\\4a+b&=0.\quad\text{(3)}\end{aligned}

\begin{aligned}b&=-4a,\\4a+2(-4a)+c&=0,\\c&=4a,\\a-(-4a)+4a&=18,\\9a&=18,\\a&=2,\\b&=-8,\\c&=8.\end{aligned}

Hence $y=2x^2-8x+8=2(x-2)^2$ . The completed-square form checks the minimum at $(2,0)$ , and $y(-1)=18$ .

Equations that become linear after substitution

Solve for the replacements, then return to the original unknowns

Solve

\begin{aligned}x^2+\sqrt y+z&=6,\\2x^2-\sqrt y+3z&=11,\\3x^2+2\sqrt y-z&=5.\end{aligned}

Let $X=x^2\geq0$ and $Y=\sqrt y\geq0$ . The system is linear in $X,Y,z$ . Subtract twice the first equation from the second, and three times the first from the third:

\begin{aligned}-3Y+z&=-1,\\-Y-4z&=-13,\\z&=3Y-1,\\-Y-4(3Y-1)&=-13,\\Y&=\frac{17}{13},\\z&=\frac{38}{13},\\X&=6-\frac{17}{13}-\frac{38}{13}\\&=\frac{23}{13}.\end{aligned}

Return to the original quantities:

\begin{aligned}x^2&=\frac{23}{13},\\x&=\pm\sqrt{\frac{23}{13}},\\\sqrt y&=\frac{17}{13},\\y&=\frac{289}{169},\\z&=\frac{38}{13}.\end{aligned}

There are two solutions, one for each sign of $x$ . The positive value of $Y$ is permitted. A negative result for $Y$ would make the original system impossible over the reals; it could not be repaired by squaring.

Decide when the solution is unique

Keep the exceptional parameter value before dividing

Discuss the solutions of

\begin{aligned}x+y+z&=6,\quad\text{(1)}\\x+2y+3z&=10,\quad\text{(2)}\\2x+3y+kz&=c.\quad\text{(3)}\end{aligned}

Add the first two equations, then compare with the third:

\begin{aligned}2x+3y+4z&=16,\\(k-4)z&=c-16.\end{aligned}

If $k\ne4$ , there is exactly one solution:

\begin{aligned}z&=\frac{c-16}{k-4},\\y+2z&=4,\\y&=4-2z,\\x&=6-y-z\\&=2+z.\end{aligned}

If $k=4$ , the reduced equation is $0=c-16$ . For $c\ne16$ , it is a contradiction, so there is no solution. For $c=16$ , the third equation repeats information already given, so there are infinitely many solutions:

\begin{aligned}(x,y,z)&=(2+t,\ 4-2t,\ t),\\t&\in\mathbb R.\end{aligned}

A zero row in the calculator does not automatically mean no solution. Distinguish $0=0$ from $0=\text{non-zero}$ .

Choose different members of the same solution family

Choose z = t:0.0

t = 0 gives (x, y, z) = (2, 4, 0). The three left sides are 6, 10 and 16.

This is the projection of the solution family onto the x-y plane. The chosen value of t supplies its third coordinate z.

Find unknown recurrence constants from consecutive terms

Use the correct index in every equation

A sequence satisfies $u_{n+1}=a2^n+bn+cu_n$ , with $u_1=1,u_2=6,u_3=16,u_4=35$ . Find $a,b,c$ , then $u_5$ .

\begin{aligned}n=1:\quad2a+b+c&=6,\\n=2:\quad4a+2b+6c&=16,\\n=3:\quad8a+3b+16c&=35.\end{aligned}

Subtract twice the first equation from the second:

\begin{aligned}4c&=4,\\c&=1,\\2a+b&=5,\\8a+3b&=19,\\(8a+3b)-3(2a+b)&=19-15,\\2a&=4,\\a&=2,\quad b=1.\end{aligned}

Use $n=4$ to generate the next term:

\begin{aligned}u_5&=2(2^4)+1(4)+1(35)\\&=71.\end{aligned}

Check the third given step: $2(2^3)+3+16=35$ . The index is the index of the input term $u_n$ , not the output term.

Before you finish

Do the equations describe the original unknowns with the right units?
Are the coefficient columns in the same order?
Did you check exceptional parameter values before division?
After substitution, did you restore all permitted roots and reject impossible replacements?

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