C2.5; E2.5 - Simultaneous linear equations and modelling

C2.5; E2.5 - Simultaneous linear equations and modelling

Two notebooks and a pen cost 11;onenotebookandthreepenscost11; one notebook and three pens cost 13. How can you find both prices when neither is given? C2.5 and E2.5 develop elimination and substitution, then use two independent contextual conditions to construct and solve a pair of equations.

Elimination

One equation with two unknowns often allows many pairs. For x+y=7x+y=7, the pairs (1,6)(1,6) and (2,5)(2,5) both work. A second condition can identify a pair that satisfies both equations at the same time.

Simultaneous solution

A solution of two simultaneous equations is a pair of values that satisfies both original equations. Finding a pair that satisfies only one equation is insufficient.

Elimination removes one unknown by adding or subtracting equations. You can add equal expressions on the left and equal expressions on the right because equality is preserved. Equal coefficients with the same sign cancel by subtraction; equal coefficients with opposite signs cancel by addition.

Worked example 1: coefficients already matched

Question. Solve 2x+y=112x+y=11 and x−y=1x-y=1.

Plan. Add the equations to eliminate yy, then substitute to find it.

(2x+y)+(x−y)=11+1add corresponding sides3x=12the y-terms cancelx=44−y=1substitute in the second equationy=3\begin{aligned} (2x+y)+(x-y)&=11+1 &&\text{add corresponding sides}\\ 3x&=12 &&\text{the }y\text{-terms cancel}\\ x&=4\\ 4-y&=1 &&\text{substitute in the second equation}\\ y&=3 \end{aligned}

Answer. x=4x=4, y=3y=3.

Check. 2(4)+3=112(4)+3=11 and 4−3=14-3=1. The pair satisfies both originals.

When coefficients do not match, multiply an entire equation by a suitable non-zero number. Multiplying just one term would change its solutions. Sometimes one equation needs scaling; sometimes both do. Use the least common multiple of the coefficients you want to match.

Worked example 2: scaling both equations

Question. Solve 2x+3y=132x+3y=13 and 3x+2y=123x+2y=12.

Plan. Make both yy-coefficients 6, then subtract the equations.

4x+6y=26first equation multiplied by 29x+6y=36second equation multiplied by 3(9x+6y)−(4x+6y)=36−265x=10x=22(2)+3y=13substitute in the first original3y=9y=3\begin{aligned} 4x+6y&=26 &&\text{first equation multiplied by 2}\\ 9x+6y&=36 &&\text{second equation multiplied by 3}\\ (9x+6y)-(4x+6y)&=36-26\\ 5x&=10\\ x&=2\\ 2(2)+3y&=13 &&\text{substitute in the first original}\\ 3y&=9\\ y&=3 \end{aligned}

Answer. x=2x=2, y=3y=3.

Check. 2(2)+3(3)=132(2)+3(3)=13, and 3(2)+2(3)=123(2)+2(3)=12.

If one coefficient already divides the other, scale only the equation that needs it. For example, coefficients 2y2y and 6y6y need the first equation multiplied by 3. Either unknown can be eliminated; choose the smaller useful multipliers.

Watch out

Multiplying 2x+3y=132x+3y=13 by 2 gives 4x+6y=264x+6y=26, not 4x+3y=264x+3y=26. When subtracting equations, subtract every term on the second line, including its sign.

Display the scaled equations so the cancellation is visible.

Substitution and checking

Substitution replaces an unknown by an expression known to be equal to it. If y=2x−1y=2x-1, every occurrence of yy can be replaced by 2x−12x-1. Put that expression into the other equation to create an equation in xx alone.

Substituting into the equation just rearranged tells you no new information: it merely repeats an equality. The second equation supplies the extra condition. This method is often convenient when an unknown is already isolated or has coefficient 1.

Worked example 3: an unknown already isolated

Question. Solve y=2x−1y=2x-1 and 3x+y=143x+y=14.

Plan. Replace yy in the second equation by 2x−12x-1, solve, then recover yy.

3x+(2x−1)=14substitute in the other equation5x−1=145x=15x=3y=2(3)−1=5\begin{aligned} 3x+(2x-1)&=14 &&\text{substitute in the other equation}\\ 5x-1&=14\\ 5x&=15\\ x&=3\\ y&=2(3)-1=5 \end{aligned}

Answer. x=3x=3, y=5y=5.

Check. The first equation gives 5=2(3)−15=2(3)-1; the second gives 3(3)+5=143(3)+5=14.

Now neither equation has an isolated unknown. A coefficient of 1 still makes one rearrangement particularly short.

Worked example 4: rearrange before substituting

Question. Solve 2y=3x−42y=3x-4 and x+y=8x+y=8.

Plan. Rearrange the second equation to y=8−xy=8-x, avoiding fractions, then substitute into the first.

y=8−x2(8−x)=3x−4replace the whole y16−2x=3x−4expand both products20=5xx=4y=8−4=4\begin{aligned} y&=8-x\\ 2(8-x)&=3x-4 &&\text{replace the whole }y\\ 16-2x&=3x-4 &&\text{expand both products}\\ 20&=5x\\ x&=4\\ y&=8-4=4 \end{aligned}

Answer. x=4x=4, y=4y=4.

Check. 2(4)=3(4)−4=82(4)=3(4)-4=8, and 4+4=84+4=8. Both original equations hold.

A substitution check must use the original equations, especially after a bracket expansion or a sign change. Elimination and substitution should agree because they preserve the same two equalities.

Watch out

If y=8−xy=8-x, then 2y=2(8−x)=16−2x2y=2(8-x)=16-2x, not 16−x16-x. Keep a substituted multi-term expression in brackets until every required multiplication has been done.

Choose the equation with the easier isolated expression.

Building two-equation models

Two unknown quantities need clear meanings before you write two equations. Keep each meaning fixed across both conditions: if nn means the price of one notebook in dollars, 2n2n is the price of two notebooks, not their number.

Use one independent condition for each equation. Doubling an equation does not create a new condition; it describes the same relationship again. Tables can keep quantities and totals aligned.

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