Simultaneous Showdown

Simultaneous Showdown

Simultaneous equations are two equations sharing the same two unknowns. The solution is the one pair of values (x, y) that makes both true at once. The method is elimination: get rid of one variable so the other can be solved. The module ends on the case nobody warns you about, where the elimination removes both variables at once and leaves something plainly false.

The words for it

Five terms. The last one is why the whole method works, and it is worth carrying in your head while you do the algebra:

Worked through

Solve x + y = 5 and x − y = 1. The y terms are +y and −y, which are opposites, so adding the two equations makes them cancel: (x + y) + (x − y) = 5 + 1, which is 2x = 6, so x = 3. Now back-substitute into either original: 3 + y = 5, so y = 2. Finally check in the OTHER equation, the one you did not substitute into: 3 − 2 = 1. ✓ That last step catches almost every arithmetic slip, and it costs about five seconds.

Add, or subtract?

Which operation cancels a variable depends on the signs of its two coefficients: • Same sign (both +2y, or both −2y) → subtract the equations. • Opposite signs (+y and −y) → add them. The memory aid is Same Subtract. And when the coefficients do not match at all, which is most of the time in an exam, multiply one or both equations through first until they do. Multiplying an entire equation by a number leaves it saying exactly the same thing, so nothing is lost: 2x + 3y = 13 and 4x + 6y = 26 are the same equation wearing different clothes.

Which operation?

You want to eliminate y from 3x + 2y = 12 and x + 2y = 8. Both y terms are +2y. What do you do, and what is left?

• Subtract the second from the first. The +2y terms cancel and you are left with 2x = 4
• Add them, because adding is how variables cancel
• Multiply the two equations together
• Nothing yet: the coefficients of x do not match, so no elimination is possible

Solve for x

Solve 3x + 2y = 12 and x + 2y = 8 by elimination. What is the value of x?

When nothing matches

Take 2x + 3y = 13 and 4x − y = 5. No coefficient matches, so something has to be multiplied first. There are two routes, both correct, and they meet at the same answer:

Solve for x

Solve 3x + 2y = 19 and x + 4y = 23. What is the value of x?

Solve for y

Same pair: 3x + 2y = 19 and x + 4y = 23. What is the value of y?

Which are true?

Select ALL THREE statements that are TRUE.

• Checking your answer in the equation you did NOT back-substitute into is a genuinely independent test of it
• Multiplying a whole equation through by a number leaves it saying exactly the same thing
• A solution is a pair of values, and both must satisfy both equations
• When you subtract one equation from another, only the variable terms are subtracted
• You can multiply the two equations together to eliminate a variable
• If you find x = 3, you have solved the simultaneous equations

Plot the solution

The lines x + 2y = 8 and x − y = 2 cross at their simultaneous solution. Solve the pair, then plot that crossing point on the grid.

When it all cancels

A student tries to solve 2x + y = 5 and 4x + 2y = 7. They multiply the first by 2 and subtract, and every variable disappears, leaving 0 = 3. What does that mean?

• There is no solution. The two lines are parallel, so they never cross, and 0 = 3 is the algebra reporting that no pair of values can satisfy both
• They have made an arithmetic mistake, since the variables should never all disappear
• The solution is x = 0 and y = 3
• There are infinitely many solutions, because any values will satisfy 0 = 3

The showdown rules

To solve linear simultaneous equations by elimination, make one variable's _____ match, multiplying an equation through first if you need to, then add or subtract to remove it. Same sign, _____. Once you have one value, back-substitute for the other, and check in the equation you did not use. On a graph the solution is the point where the two lines _____, so a pair with no solution appears as two _____ lines.