Systems of Equations Practice Questions

Since the first equation is already solved for \(y\), substitute \(2x + 1\) for \(y\) in the second equation:

\(3x + (2x + 1) = 11\)

\(5x + 1 = 11\)

\(5x = 10\)

\(x = 2\)

Now substitute \(x = 2\) back into the first equation to find \(y\):

\(y = 2(2) + 1 = 5\)

The solution is \((2, 5)\). You can verify by checking both equations:

\(3(2) + 5 = 11\)

Notice that the \(y\)-coefficients are opposites (\(+3y\) and \(-3y\)). Add the two equations to eliminate \(y\):

\((2x + 3y) + (4x – 3y) = 12 + 6\)

\(6x = 18\)

\(x = 3\)

Substitute \(x = 3\) into the first equation:

\(2(3) + 3y = 12 \implies 3y = 6 \implies y = 2\)

The solution is \((3, 2)\). Verify with the second equation:

\(4(3) – 3(2) = 12 – 6 = 6\)

Both equations are in slope-intercept form \((y = mx + b)\). Compare the slopes and \(y\)-intercepts:

• Equation 1: slope = 3, \(y\)-intercept = 2
• Equation 2: slope = 3, \(y\)-intercept = −4

The slopes are equal but the \(y\)-intercepts are different, which means the lines are parallel. Parallel lines never intersect, so the system has no solution.

A system with no solution is called inconsistent. If the slopes were different, the lines would intersect at exactly one point. If both the slopes and \(y\)-intercepts were equal, the lines would overlap and there would be infinitely many solutions.

Divide every term in the first equation by 3:

\(\dfrac{3x – 6y}{3} = \dfrac{12}{3} \implies x – 2y = 4\)

This is identical to the second equation. Since both equations represent the same line, every point on that line is a solution. The system has infinitely many solutions.

A system like this is called dependent. Whenever one equation is a scalar multiple of the other, the system is dependent and has infinitely many solutions.

Solve the first equation for \(x\):

\(x = 8 – 2y\)

Substitute into the second equation:

\(3(8 – 2y) – y = 3\)

\(24 – 6y – y = 3\)

\(24 – 7y = 3\)

\(-7y = -21\)

\(y = 3\)

Substitute back: \(x = 8 – 2(3) = 2\). The solution is \((2, 3)\).

Neither variable cancels immediately, so multiply each equation by a value that creates opposite coefficients. To eliminate \(y\), multiply the first equation by 3 and the second by 2:

\(3(3x + 2y) = 3(16) \implies 9x + 6y = 48\)

\(2(5x – 3y) = 2(14) \implies 10x – 6y = 28\)

Now add the equations to eliminate \(y\):

\(19x = 76 \implies x = 4\)

Substitute \(x = 4\) into the first original equation:

\(3(4) + 2y = 16 \implies 2y = 4 \implies y = 2\)

The solution is \((4, 2)\).

Translate the problem into a system of equations. Let \(x\) be the larger number and \(y\) be the smaller number:

\(\begin{cases} x + y = 25 \\ x – y = 7 \end{cases}\)

Add the two equations to eliminate \(y\):

\(2x = 32 \implies x = 16\)

Substitute back: \(16 + y = 25 \implies y = 9\).

The two numbers are 16 and 9.

Let \(a\) represent the number of adult tickets and \(c\) represent the number of child tickets. Set up the system:

\(\begin{cases} a + c = 12 \\ 8a + 5c = 81 \end{cases}\)

Solve the first equation for \(c\): \(c = 12 – a\). Substitute into the second equation:

\(8a + 5(12 – a) = 81\)

\(8a + 60 – 5a = 81\)

\(3a = 21\)

\(a = 7\)

The group purchased 7 adult tickets (and 5 child tickets).

The solution to a system of equations must satisfy both equations simultaneously. Test \((3, 1)\) in each equation:

• Equation 1: \(2(3) + 1 = 6 + 1 = 7\)
• Equation 2: \(3 – 1 = 2\)

Both equations are satisfied, so \((3, 1)\) is the solution. You can verify that the other choices fail at least one equation. For example, \((4, -1)\) satisfies the first equation (\(2(4) + (-1) = 7\)) but not the second (\(4 – (-1) = 5 \neq 2\)).

A solution to a system of inequalities must satisfy every inequality. Test \((1, 3)\):

• Inequality 1: \(3 \gt 2(1) – 1 = 1\) → \(3 \gt 1\)
• Inequality 2: \(3 \leq -(1) + 5 = 4\) → \(3 \leq 4\)

Both inequalities are satisfied. Checking the other choices:

• Choice B: \((3, 1)\) → \(1 \gt 2(3) – 1 = 5\)? No.
• Choice C: \((0, -2)\) → \(-2 \gt 2(0) – 1 = -1\)? No.
• Choice D: \((4, 2)\) → \(2 \gt 2(4) – 1 = 7\)? No.

Since both equations equal \(y\), set the right sides equal to each other:

\(x^2 = x + 2\)

Rearrange into standard form and factor:

\(x^2 – x – 2 = 0\)

\((x – 2)(x + 1) = 0\)

So \(x = 2\) or \(x = -1\). Find the corresponding \(y\)-values using \(y = x + 2\):

• \(x = 2\): \(y = 2 + 2 = 4\) → \((2, 4)\)
• \(x = -1\): \(y = -1 + 2 = 1\) → \((-1, 1)\)

The system has two solutions. Graphically, this means the parabola and the line intersect at two points.

Let \(x\) be the liters of 20% solution and \(y\) be the liters of 50% solution. Set up two equations — one for total volume and one for total acid:

\(\begin{cases} x + y = 12 \\ 0.20x + 0.50y = 0.30(12) \end{cases}\)

Simplify the second equation: \(0.20x + 0.50y = 3.6\). Solve the first equation for \(y\):

\(y = 12 – x\)

Substitute:

\(0.20x + 0.50(12 – x) = 3.6\)

\(0.20x + 6 – 0.50x = 3.6\)

\(-0.30x = -2.4\)

\(x = 8\)

The chemist needs 8 liters of the 20% solution (and 4 liters of the 50% solution).

Translate each sentence into an equation:

• “The length is 3 more than twice the width.”
  \(l = 2w + 3\)
• “The perimeter is 42 cm.”
  \(2l + 2w = 42\)

Choice A uses \(l + w = 42\), which is the semi-perimeter, not the full perimeter. Choice C reverses the relationship between length and width. Choice D uses the area formula (\(lw\)) instead of the perimeter formula.

One approach is to add both equations together:

\((x + 2y) + (2x + y) = 12 + 9\)

\(3x + 3y = 21\)

Factor out the 3:

\(3(x + y) = 21 \implies x + y = 7\)

This shortcut avoids solving for \(x\) and \(y\) individually. When a question asks for a combination of variables rather than individual values, look for ways to manipulate the equations directly to get that combination.

Let \(m\) represent Maria’s current age and \(d\) represent her daughter’s current age. Translate each statement into an equation:

• “Maria is 4 times as old as her daughter.”
  \(m = 4d\)
• “In 10 years, Maria will be twice as old.”
  \(m + 10 = 2(d + 10)\)

Substitute \(m = 4d\) into the second equation:

\(4d + 10 = 2(d + 10)\)

\(4d + 10 = 2d + 20\)

\(2d = 10\)

\(d = 5\)

The daughter is 5 years old (and Maria is 20).