Back to Tutor Intake Assessment: Solve systems of two linear equations in two variables algebraically

8.EE.C.8.b Tutor Intake -- Solving Systems Algebraically

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Grade 8·8 problems·~9 min·Common Core Math - Grade 8·standard·8-ee-c-8b

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Concepts

A system has the equations $3x + 2y = 5$ and $3x + 2y = 6$ . By
inspection (without doing algebra), how many solutions does this
system have?

No solution — the same expression $3x + 2y$ cannot equal both 5 and 6 at once.

One solution — substitute the expression from the first equation into the second equation and solve.

Infinitely many solutions, since both equations have the same left side.

It cannot be determined without graphing.

A student solves a system using elimination and arrives at
$x = -3/4$ . Their partner says the negative fraction means the
system has "no solution." Is the partner correct?

No — a solution can be any real number, including negative numbers and fractions. Check by substituting $x = -3/4$ into both original equations.

Yes — a fractional or negative result always signals no solution.

Yes, but only because the value is negative, not because it's a fraction.

It depends on whether the y-value also turns out to be a fraction.

Procedures

Solve the system $y = 2x + 1$ and $3x + y = 11$ using
substitution: substitute $(2x + 1)$ for $y$ in the SECOND
equation, $3x + y = 11$ , and solve for $x$ .

Solve the system $2x + 3y = 12$ and $2x - y = 4$ using
elimination: subtract the second equation from the first to
eliminate $x$ , then solve for $y$ .

Solve the system $x + 2y = 7$ and $3x - y = 7$ using
elimination. First multiply the first equation by 3 to match the
x-coefficients, giving $3x + 6y = 21$ . Then subtract the second
equation from this result and solve for $y$ .

A student solves the system $2x + 3y = 12$ and $2x - y = 4$ and
finds $y = 2$ . They stop there and report " $y = 2$ " as the final
answer. Substitute $y = 2$ into $2x - y = 4$ and find the value
of $x$ that completes the ordered-pair solution.

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