Systems of equations: Algebra 1 Regents Practice

What is the solution to the system of equations $y = 2x - 3$ and $x + y = 9$?

• (A) $(4,\ 11)$
• (B) $(5,\ 4)$
• (C) $(4,\ 5)$ ✓
• (D) $(3,\ 3)$

In substitution, replace one variable using one equation, solve the resulting single-variable equation, then back-substitute to find the other coordinate.

What is the solution to the system of equations $x + y = 7$ and $x - y = 3$?

• (A) $(10,\ -3)$
• (B) $(5,\ -2)$
• (C) $(2,\ 5)$
• (D) $(5,\ 2)$ ✓

Elimination adds or subtracts the equations so one variable cancels; add when a variable already has opposite coefficients.

A tour bus seats at most $48$ passengers. An adult ticket is $\$18$ and a child ticket is $\$12$, and the company must collect at least $\$650$ to make a profit. If $a$ is the number of adult tickets and $c$ is the number of child tickets, which system models a profitable trip?

• (A) $a + c < 48$ and $18a + 12c < 650$
• (B) $a + c \le 48$ and $18a + 12c \ge 650$ ✓
• (C) $a + c \le 48$ and $18a + 12c \le 650$
• (D) $a + c < 48$ and $18a + 12c > 650$

Translate the words: 'at most' is ≤, 'at least' is ≥. Each real-world limit becomes one inequality, and the answer is the system of both.

A theater sells adult tickets for $\$8$ and child tickets for $\$5$. On one night $200$ tickets were sold for a total of $\$1360$. How many adult tickets were sold?

• (A) $100$
• (B) $170$
• (C) $80$
• (D) $120$ ✓

Translate the two conditions into two equations, substitute to solve, and be sure to report the variable the question actually asks for.

How many solutions does the system of equations $6x - 2y = 10$ and $y = 3x - 5$ have?

• (A) Infinitely many solutions ✓
• (B) Exactly two solutions
• (C) No solution
• (D) Exactly one solution, $(0, -5)$

If two linear equations simplify to the same line, the system is dependent with infinitely many solutions; equal slopes but different intercepts give none, and different slopes give exactly one.

What is the solution to the system of equations $y = 2x + 1$ and $y = -x + 7$?

• (A) $(6, 13)$
• (B) $(2, 5)$ ✓
• (C) $(-2, -3)$
• (D) $(1, 3)$

Since both equations are solved for $y$, set them equal and solve for $x$. This gives $x = 2$, and substituting into either equation yields $y = 5$. The distractors satisfy only the first equation, which happens when a student stops before checking the point in both equations.

Solve the following system of equations algebraically: $y = 2x + 1$ and $3x + 2y = 16$

Because one equation is already solved for $y$, substitution is the efficient method. Replacing $y$ produces a single linear equation in $x$; solving it and back-substituting yields the ordered pair. Checking $3(2)+2(5)=16$ confirms the solution.

At a school concession stand, each hot dog sells for 3 dollars and each bottle of water sells for 2 dollars. During one game, a total of 250 items were sold, and the total amount of money collected was 630 dollars. Write a system of equations that models this situation, and solve it algebraically to determine how many hot dogs and how many bottles of water were sold.

Let $h$ be hot dogs and $w$ be bottles of water. One equation counts the items and the other counts the money. Solving by substitution gives $h = 130$ and $w = 120$, and a check confirms $130 + 120 = 250$ items and $3(130) + 2(120) = 630$ dollars.

What is the solution to the system of equations $y = 2x - 1$ and $x + y = 8$?

• (A) $(3, 5)$ ✓
• (B) $(2, 3)$
• (C) $(5, 3)$
• (D) $(3, 7)$

Substituting the first equation into the second gives $3x - 1 = 8$, so $x = 3$ and $y = 5$. Check: $3 + 5 = 8$ and $2(3) - 1 = 5$, both true. The ordered pair must be written as $(x, y)$, so $(3, 5)$, not $(5, 3)$.

What is the value of $x$ in the solution to the system of equations below? $$2x + 3y = 12$$ $$x - y = 1$$

• (A) $-3$
• (B) $2$
• (C) $3$ ✓
• (D) $5$

Solving the second equation for $x$ and substituting into the first gives $5y + 2 = 12$, so $y = 2$ and $x = 3$. Choosing $2$ reports the $y$-value instead of the $x$-value, a very common mix-up, while the other distractors come from arithmetic slips in the substitution.

A school sold a total of $120$ tickets to a play. Adult tickets cost $\$9$ each and student tickets cost $\$6$ each. The total amount collected was $\$855$. Write a system of equations that models this situation, and use it to determine algebraically the number of adult tickets and the number of student tickets sold.

One equation counts tickets and the other counts money. Eliminating $s$ by scaling the count equation isolates the adult tickets, giving $a=45$, and substitution gives $s=75$. Both values satisfy the total of $\$855$.

What is the solution to the system of equations below? $$y = 2x - 3$$ $$3x + y = 12$$

• (A) $(3, 3)$ ✓
• (B) $(5, 7)$
• (C) $(3, -3)$
• (D) $(1, -1)$

Since the first equation gives $y$ in terms of $x$, substitution is efficient. Replacing $y$ in $3x + y = 12$ gives $5x - 3 = 12$, so $x = 3$ and $y = 2(3) - 3 = 3$. The point $(3, 3)$ satisfies both equations, which is the definition of a system's solution.

Given the system of equations $y = 2x + 1$ and $3x + y = 16$, what is the value of $y$?

• (A) $3$
• (B) $7$ ✓
• (C) $13$
• (D) $5$

Substitution replaces $y$ in the second equation with $2x + 1$, giving $5x + 1 = 16$, so $x = 3$. The question asks for $y$, so substitute back: $y = 2(3) + 1 = 7$. Answering $3$ is the classic error of reporting $x$ instead of $y$.

At a school store, 3 T-shirts and 2 hats cost $\$46$. At the same prices, 2 T-shirts and 4 hats cost $\$44$. Determine algebraically the cost of one T-shirt and the cost of one hat.

Writing one equation for each purchase gives a system of two linear equations. Multiplying the first equation by 2 makes the hat coefficients equal, so subtracting eliminates $h$ and gives $s=12$; substituting back gives $h=5$. Both original equations check: $3(12)+2(5)=46$ and $2(12)+4(5)=44$.

A school theater sold 12 tickets to a play for a total of $\$76$. Adult tickets cost $\$9$ each and child tickets cost $\$5$ each. Write a system of equations that models this situation, where $a$ is the number of adult tickets and $c$ is the number of child tickets. Use your system to determine algebraically the number of each type of ticket sold.

The two conditions, total number of tickets and total revenue, give a linear system. Elimination removes one variable: multiplying $a + c = 12$ by 5 and subtracting from the money equation yields $4a = 16$, so $a = 4$ and $c = 8$, which satisfies both original equations.

What is the solution to the system of equations below? $$y = 2x - 3$$ $$3x + 2y = 15$$

• (A) $(5, 0)$
• (B) $(2, 1)$
• (C) $(1, -1)$
• (D) $(3, 3)$ ✓

Substituting the expression for $y$ into the second equation gives $7x = 21$, so $x = 3$ and $y = 3$. The solution of a system must satisfy both equations; each distractor satisfies only one of the two equations, which is why checking in both is essential.

What is the value of $y$ in the solution of the system of equations below? $$3x + 2y = 16$$ $$x - 2y = 0$$

• (A) $2$ ✓
• (B) $-2$
• (C) $4$
• (D) $8$

Adding the two equations eliminates $y$ because the $y$ coefficients are opposites, giving $4x = 16$, so $x = 4$. Substituting back into $x - 2y = 0$ gives $y = 2$. The distractor $4$ is the value of $x$, a common mix-up when a system asks for one specific variable.

A movie theater sold a total of 12 tickets for one showing. Adult tickets cost $\$9$ each and child tickets cost $\$5$ each. The total amount collected was $\$80$. Write a system of equations that models this situation, and determine algebraically the number of adult tickets and the number of child tickets sold.

One equation counts tickets and the other counts money. Eliminating $c$ by subtracting a scaled copy of the ticket equation leaves $4a = 20$, so $a = 5$ and $c = 7$, which satisfies both the ticket total and the money total.

At a school play, two adult tickets and three student tickets cost a total of $\$40$. One adult ticket and one student ticket cost a total of $\$15$. Write a system of equations that models this situation, and determine algebraically the price of one adult ticket and the price of one student ticket.

Defining a variable for each ticket price gives a system of two linear equations. Multiplying the second equation by 2 lines up the a terms, so elimination isolates s. Substituting back gives a, and the pair $(a, s) = (5, 10)$ satisfies both original equations.

At a coffee shop, $4$ lattes and $2$ donuts cost $\$15.50$ one day; the next day, $3$ lattes and $5$ donuts cost $\$18.10$ (tax included). Let $x$ be the cost of one latte and $y$ the cost of one donut. Write a system of equations, decide whether one latte costs $\$2.75$ and one donut $\$2.25$, and find the exact cost of each.

A guess must satisfy every equation in the system, checking only one isn't enough. Set up one equation per scenario, then solve by elimination or substitution.

Solve the system $y = x^2 - 3x - 6$ and $y = x - 1$ graphically, and state the coordinates of all solutions.

The solutions of a system are the intersection points. Graph both, or set the expressions equal and solve the quadratic, then find each matching y.

Solve the system algebraically for all $x$ and $y$: $\;y=x^{2}+9x+4\;$ and $\;y-2x=-6$.

Substitute the linear expression for y, solve the resulting quadratic, then back-substitute for each y.

What are the solutions to the system of equations $y = x^2 - 2x - 3$ and $y = x - 3$?

• (A) $(3, 0)$ and $(-1, -4)$
• (B) $(3, 0)$ only
• (C) $(0, -3)$ and $(3, 0)$ ✓
• (D) $(-3, 0)$ and $(0, 3)$

For a linear-quadratic system, substitute the line into the parabola and solve the resulting quadratic; each real $x$-value produces one intersection point.

At a snack stand, hot dogs cost $\$4$ and drinks cost $\$2$. Jordan buys $10$ items for a total of $\$28$. How many hot dogs did Jordan buy?

• (A) $7$
• (B) $5$
• (C) $6$
• (D) $4$ ✓

Write one equation for the number of items and one for the total cost, then substitute; the count equation is essential and cannot be discarded.