➕ Algebra 1 · Systems of Equations
Solving Systems by Substitution: true or false
Decide fast, then read the reason. Wrong answers are the best teachers, so every statement explains itself.
All statements and answers
Substitution replaces a variable in one equation with an expression for that variable taken from the other equation.
✓ True. At a solution the two are equal, so the swap keeps the equation true and leaves only one variable.
Substituting y = 7 − 2x into 3x + 4y = 5 gives 3x + 4 · 7 − 2x = 5.
✗ False. The expression needs parentheses: 3x + 4(7 − 2x) = 5. Without them the 4 multiplies only the 7.
If every variable cancels and you are left with 7 = 7, the system has infinitely many solutions.
✓ True. A true leftover means the new equation is true for every value of the variable, so the two equations describe the same line.
If every variable cancels and you are left with 0 = 5, the solution is x = 0 and y = 5.
✗ False. 0 = 5 is false for every x and y, so the system has no solution. The numbers 0 and 5 are not coordinates.
In a system with exactly one solution, once you have the correct value of one variable you can find the other by putting it into any original equation that contains the other variable.
✓ True. The solution pair makes both equations true, so any equation that contains the other variable gives its value.
In substitution you must always solve for x first.
✗ False. You can isolate either variable in either equation. Choose the one that gives the least work, usually a variable with coefficient 1 or −1.
You can use substitution even when no variable has coefficient 1 or −1, but isolating a variable then often produces fractions.
✓ True. For example, 2x + 3y = 6 gives y = 2 − (2/3)x. It is still correct, just messier. (Not always: in 2x + 4y = 10, x = 5 − 2y has no fractions, because 2 divides every term.)
If you substitute an expression into the same equation it came from, you will find the solution.
✗ False. You get a statement that is always true, like y = y, and you learn nothing. Substitute into the other equation.
The pair (−1, 4) is the solution of x + y = 3 and x = 2y + 5.
✗ False. In the first equation, −1 + 4 = 3 is true. In the second, 2(4) + 5 = 13, not −1. The pair fits only one equation, so it is not the solution.
If every variable cancels and you are left with 2 = 11, you must have made an arithmetic mistake.
✗ False. Not necessarily. Recheck your steps, but if they are right, a false leftover means the system has no solution: the lines are parallel.
If x = −3, then 2x − 5 equals 2(−3) − 5 = −11.
✓ True. The parentheses keep the sign: 2(−3) = −6, and −6 − 5 = −11.
The system y = 4x + 1 and y = 4x + 7 has no solution, because substituting gives 4x + 1 = 4x + 7, which simplifies to the false statement 1 = 7.
✓ True. Subtract 4x from both sides to get 1 = 7. No value of x makes that true, so the lines (same slope, different y-intercepts) never meet.