14. Linear equations and systems
Count equations vs unknowns. Special cases: dependent and inconsistent systems.
Core ideas
- Count equations vs unknowns: n independent equations usually pin down n unknowns, but only if they are truly independent.
- If one equation is a multiple of another, they are dependent: infinitely many solutions, you cannot solve for individual variables.
- Same left side, different right side means inconsistent: no solution at all.
- On Data Sufficiency, always ask whether the second equation adds new information or just restates the first.
Worked example 1
If 2x + 3y = 12 and 4x + 6y = 24, what is the value of x?
Show solution
It is tempting to say two equations, two unknowns, so x is solvable. But the second equation is exactly 2 times the first, so it carries no new information. There is really only one equation with two unknowns, which has infinitely many solutions: for instance x = 0, y = 4 works, and so does x = 3, y = 2. The value of x cannot be determined. The move is to check for a multiple before trusting the equation count.
Worked example 2
If 3x + 2y = 19 and x - y = 3, what is x?
Show solution
Here the equations are genuinely independent, so a unique solution exists. From the second equation, x = y + 3. Substitute into the first: 3(y + 3) + 2y = 19, so 5y + 9 = 19, giving y = 2. Then x = 2 + 3 = 5. Check: 3(5) + 2(2) = 19 and 5 - 2 = 3, both true.
Practice set
Question 1
If 2x + 3y = 12 and x = 3, what is the value of x + y?
Substitute the known value of x first, then read off what the question actually asks.
Substituting x = 3 into 2x + 3y = 12 gives 6 + 3y = 12, so 3y = 6 and y = 2. Then x + y = 3 + 2 = 5. Answering 2 stops at y instead of finishing with x + y.
Question 2
The system of equations 2x - 5y = 7 and 6x - 15y = k has at least one solution. What is the value of k?
Compare the left sides first; one is an exact multiple of the other, which forces the right sides to match.
The left side of the second equation is exactly 3 times the left side of the first, since 6 = 3(2) and -15 = 3(-5). So the second equation can only be consistent with the first if its right side is also 3 times 7, that is k = 21. For any other k the two lines are parallel and distinct, giving no solution; at k = 21 they coincide and there are infinitely many solutions.
Question 3
For what value of a does the system ax + 3y = 9 and 2x + y = 5 have no solution?
No solution means proportional coefficients but a constant term that breaks the proportion.
The system has no solution when the lines are parallel but distinct: a/2 = 3/1 but 9/5 does not equal 3. From a/2 = 3 we get a = 6. Checking: with a = 6 the first equation is 6x + 3y = 9, which simplifies to 2x + y = 3, contradicting 2x + y = 5. So the lines are parallel and never meet. If the constants had also matched, the answer would have been infinitely many solutions instead.
Question 4
At a shop, 3 pens and 2 pencils cost $12.00, and 5 pens and 2 pencils cost $18.00. What is the cost of 1 pen and 1 pencil together?
Subtracting the two totals eliminates the pencils in one step.
Subtracting the first equation from the second gives 2 pens = $6.00, so a pen costs $3.00. Then 3(3.00) + 2 pencils = 12.00 gives 2 pencils = $3.00, so a pencil costs $1.50. One pen and one pencil cost 3.00 + 1.50 = $4.50. The answer $3.00 stops at the pen price.