StudyPebble

SAT Systems of Equations With No Solution

Some system questions never ask you to solve anything. Instead they hand you two equations with an unknown constant in one of them and ask which value of that constant gives the system no solution. It is a graph question in disguise: no solution means the two lines are parallel and never cross.

The rule

A system of two linear equations has no solution when the two lines have the same slope but different y-intercepts.

Two linear equations describe two lines. If the lines cross once, the system has exactly one solution. If they lie on top of each other, every point works and there are infinitely many solutions. If they are parallel and separate, they never touch, so there is no solution at all. Every no-solution question is really asking you to force the parallel case.

Parallel means equal slopes. The fastest route is to put both equations in slope-intercept form and set the slopes equal. Take the system 3x plus 2y equals 8 and 9x plus ky equals 5. Solving the first for y gives y equals negative three halves x plus 4, so the slope is negative three halves. Solving the second gives y equals negative 9 over k times x plus 5 over k, so the slope is negative 9 over k. Setting negative three halves equal to negative 9 over k gives 3k equals 18, so k equals 6.

There is a faster version once you trust it. If you write both equations in the form ax plus by equals c, the lines are parallel exactly when the coefficient ratios match but the constant ratio does not. In the example, 9 divided by 3 is 3, so for the y coefficients to scale the same way you need k divided by 2 to equal 3, which again gives k equals 6. This coefficient-ratio move avoids fractions entirely and is usually the cleaner scratch work.

The last step is the one people skip. After you find the constant, check that the constants on the right side do not scale by the same factor. With k equal to 6 the second equation becomes 9x plus 6y equals 5. Tripling the first equation gives 9x plus 6y equals 24, and 24 is not 5, so the lines really are separate. If the right sides had matched, the two equations would describe the same line and the answer would be infinitely many solutions instead of none.

If the question gives you numbers with no unknown constant and asks how many solutions the system has, run the same comparison in reverse. Compare the two slopes first. Different slopes means one solution and you are done. Same slopes sends you to the constants, which split parallel from identical.

The trap

Finding the constant that makes the coefficients proportional and stopping there, without checking the constant terms, so an infinitely-many-solutions system gets called no solution.

Try it

Try one
Real question
10203040506020406080Miles drivenTotal charge (dollars)Company ACompany B(40, 60)

The graph shows the total charge, in dollars, for renting a van for m miles from two companies. What is the total charge, in dollars, at the mileage where the two companies charge the same amount?

A jar holds nickels and dimes in the ratio 3 to 4. The coins are worth $5.50 in all. How many dimes are in the jar?

  • A) 10
  • B) 30
  • C) 40 ✓
  • D) 70

Why: Write the counts as 3k nickels and 4k dimes. Their value is 0.05(3k) + 0.10(4k) = 0.55k = 5.50, so k = 10 and there are 40 dimes. Answering 70 gives the total number of coins. Answering 10 means you reported the ratio multiplier rather than the number of dimes. Answering 30 means you reported the number of nickels instead of dimes. Answering 70 means you reported the total number of coins.

Practice more of these free

Questions students ask

How do I know if a system has no solution or infinitely many?

Both cases start the same way, with the two lines having equal slopes. Then compare the constant terms. If one whole equation is an exact multiple of the other, including the number on the right side, the lines are identical and there are infinitely many solutions. If only the variable side scales, the lines are parallel and separate, so there is no solution.

Can I just try the answer choices?

Yes, and it works well here. Substitute each choice for the constant, then check whether the resulting equations are multiples of each other on the variable side only. Start with a choice that makes the coefficients look proportional, since that is usually visible at a glance.

Guides that cover this

More on systems of linear equations