FA16–FA17, HA19, HA21 - Simultaneous linear equations

FA16–FA17, HA19, HA21 - Simultaneous linear equations

One equation can allow many pairs of values; two equations can identify the pair that works in both. FA16 and the linear part of HA19 cover elimination, substitution and graphical solutions. FA17 and HA21 turn linked real information into equations and interpret the answers.

Elimination

Two notebooks and a pen cost a fixed amount, but many price pairs could fit that total. A second purchase gives another constraint. A simultaneous solution is one set of values satisfying both equations at the same time.

Elimination combines equations so that one unknown disappears. Align like terms. If the chosen unknown has equal coefficients, subtract; if its coefficients are opposites, add. You may multiply every term of an equation by the same non-zero number without changing its solutions.

Worked example 1: matching coefficients

Question. Solve 3x+2y=183x+2y=18 and x+2y=10x+2y=10.

Plan. Subtract the second equation from the first to eliminate yy.

(3x+2y)−(x+2y)=18−102x=8x=44+2y=10substitute into the second equationy=3\begin{aligned} (3x+2y)-(x+2y)&=18-10\\ 2x&=8\\ x&=4\\ 4+2y&=10 &&\text{substitute into the second equation}\\ y&=3 \end{aligned}

Answer. x=4, y=3x=4,\ y=3.

Check. 3(4)+2(3)=183(4)+2(3)=18 and 4+2(3)=104+2(3)=10. Both original totals match.

Now the coefficients do not match yet. Create opposite coefficients before adding.

Worked example 2: scaling before elimination

Question. Solve 2x+3y=192x+3y=19 and 3x−2y=93x-2y=9.

Plan. Multiply the first equation by two and the second by three to make +6y+6y and −6y-6y.

4x+6y=38first equation multiplied by 29x−6y=27second equation multiplied by 313x=65add the equationsx=52(5)+3y=19⇒y=3\begin{aligned} 4x+6y&=38 &&\text{first equation multiplied by 2}\\ 9x-6y&=27 &&\text{second equation multiplied by 3}\\ 13x&=65 &&\text{add the equations}\\ x&=5\\ 2(5)+3y&=19\quad\Rightarrow\quad y=3 \end{aligned}

Answer. x=5, y=3x=5,\ y=3.

Check. 2(5)+3(3)=192(5)+3(3)=19 and 3(5)−2(3)=93(5)-2(3)=9.

Subtraction applies to the entire equation, including negative terms.

Watch out

Multiplying 2x+3y=192x+3y=19 by two gives 4x+6y=384x+6y=38. Leaving the right side as 1919 changes the equation and generally loses the original solution.

Eliminate opposite coefficients in a fresh pair.

Substitution

When one equation already gives an unknown on its own, substitute that expression into the other equation. The result has only one unknown and can be solved using ordinary linear-equation methods.

Substitution in simultaneous equations

If y=2x+1y=2x+1, every occurrence of yy in the other equation can be replaced by the whole expression 2x+12x+1. Brackets keep that expression together when it is multiplied or subtracted.

Start where the equation already makes yy the subject.

Worked example 3: replacing y

Question. Solve y=2x+1y=2x+1 and 3x+y=163x+y=16.

Plan. Replace yy in the second equation by 2x+12x+1.

3x+(2x+1)=165x=15x=3y=2(3)+1=7\begin{aligned} 3x+(2x+1)&=16\\ 5x&=15\\ x&=3\\ y&=2(3)+1=7 \end{aligned}

Answer. x=3, y=7x=3,\ y=7.

Check. 7=2(3)+17=2(3)+1 and 3(3)+7=163(3)+7=16.

Change which variable is isolated. Substitution is equally useful for xx.

Worked example 4: replacing x

Question. Solve x=3y−2x=3y-2 and 2x+y=172x+y=17.

Plan. Replace xx by the whole expression 3y−23y-2.

2(3y−2)+y=176y−4+y=177y=21y=3x=3(3)−2=7\begin{aligned} 2(3y-2)+y&=17\\ 6y-4+y&=17\\ 7y&=21\\ y&=3\\ x&=3(3)-2=7 \end{aligned}

Answer. x=7, y=3x=7,\ y=3.

Check. 7=3(3)−27=3(3)-2 and 2(7)+3=172(7)+3=17.

Both methods solve the same problem. Prefer substitution when one variable is easy to isolate; prefer elimination when coefficients already match or can be made to match simply.

Watch out

Replacing 2x2x with 6y−26y-2 in the second example misses a multiplication. It must be 2(3y−2)=6y−42(3y-2)=6y-4: the two multiplies both terms.

Continue this lesson.

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