Question
3x + 4y = -23
2y – x = -19
What is the solution (x,y) to the system of equations above?
A. (-5,-2)
B. (3,-8)
C. (4,-6)
D. (9,-6)
Solution
The solution is (3, -8). This is one of those systems of equations SAT problems where the fastest path is substitution. The second equation is already nearly solved for x, so rearranging it and plugging it into the first equation collapses two unknowns into one clean step.
Bottom Line
(x, y) = (3, -8)
Rearranging the second equation gives x = 2y + 19. Substituting into the first equation produces a single-variable equation that solves to y = -8, and then x = 3.
We created a step-by-step video walkthrough of this system of equations SAT problem. Watch how we isolate a variable in the simpler equation, substitute into the other, and then back-solve to find both unknowns.
Step-by-Step Explanation
Like most systems of linear equations problems, this one follows a clean four-step method. Each step sets up the next.
Step 1. Choose the Equation to Isolate a Variable
Before doing any algebra, identify which equation and which variable will be easiest to isolate. The second equation, 2y minus x equals negative 19, contains x with a coefficient of negative 1. Moving it requires only one step, which makes it the better starting point. Choosing the simpler equation first is a core strategy for any equation model problem.
- 2y – x = -19
- x = 2y + 19
Step 2. Substitute into the Other Equation
Replace x in the first equation with the expression found in Step 1. Every instance of x becomes 2y + 19. This is the defining move in substitution and the reason it works: we trade two unknowns for one.
- 3x + 4y = -23
- 3(2y + 19) + 4y = -23
- 6y + 57 + 4y = -23
- 10y + 57 = -23
- 10y = -80
- y = -8
Step 3. Back-Substitute to Find the Second Variable
With y known, substitute it back into the isolated expression from Step 1 to find x. This is called back-substitution and it closes the loop on the two-variable system.
- x = 2y + 19
- x = 2(-8) + 19
- x = -16 + 19
- x = 3
Step 4. Verify the Solution in Both Original Equations
Always check the ordered pair in both original equations. A solution that works in only one equation is not a solution to the system.
- Check equation 1: 3(3) + 4(-8) = 9 – 32 = -23. Correct
- Check equation 2: 2(-8) – 3 = -16 – 3 = -19. Correct
The solution is (3, -8), which matches choice B. The core insight for any equation of a line word problems situation like this is to look for the variable with the simplest coefficient first. That single choice drives every step that follows.
Now try these two practice problems using the same method. Work through each one before checking the answer.
Practice Problem 1
5x + 2y = 16
x – y = 1
Solution
From the second equation, x = y + 1. Substitute into the first – 5(y + 1) + 2y = 16, which gives 7y + 5 = 16, so y = 1. Then x = 1 + 1 = 2.
Final answer – x = 2, y = 1
Practice Problem 2
3x – y = 7
x + 2y = 7
Solution
From the first equation, y = 3x – 7. Substitute into the second – x + 2(3x – 7) = 7, which gives 7x – 14 = 7, so x = 3. Then y = 9 – 7 = 2.
Final answer – x = 3, y = 2
Frequently Asked Questions
Solving a system of equations means finding the values of the variables that satisfy all equations in the system at the same time. For a two-equation, two-variable system like this one, the solution is an ordered pair (x, y) that makes both equations true simultaneously. In this problem that pair is (3, -8).
Use substitution when one equation already has a variable with a coefficient of 1 or -1, because isolating it takes only one algebraic step. In this problem, x in the second equation has a coefficient of -1, making substitution the faster choice. Elimination works better when both equations have matching or easily scaled coefficients. Knowing which method to reach for is one of the key skills for systems of equations SAT questions.
Checking in both equations confirms the solution is valid for the entire system, not just one equation. Arithmetic errors during substitution can produce a value that satisfies one equation but not the other. A quick verification step catches those errors before they cost you the question.
On the SAT, word problems using systems of equations are often presented with the equations already written out, as in this problem, or with a short scenario you must translate into two equations yourself. Either way, the solving method is the same: identify what substitution or elimination step simplifies the system fastest, carry it through carefully, and verify the result. These problems consistently appear in the heart of the algebra section.
Define your unknowns, set up one equation per relationship given in the problem, then use substitution or elimination to reduce the system to a single-variable equation. Solve for that variable, back-substitute to find the other, and verify both answers in the original equations. That sequence handles any word problems on linear equations you will encounter.
Equation models translate real-world constraints into mathematical language. A system of linear equations can represent anything from pricing and quantities to speeds and distances. The algebraic technique is identical regardless of context: set up the equations, solve the system, and interpret the result in the original scenario. This is exactly why equation models appear so frequently in standardized testing.
A system has one solution when the two lines intersect at a single point, as in this problem. It has no solution when the lines are parallel and never meet, which happens when the equations are inconsistent. It has infinitely many solutions when both equations describe the same line. For any equation of a line word problems context, checking whether the slopes and intercepts match will tell you which case you are dealing with.
A basic equation solver gives you the answer. Think10x.ai gives you a video that teaches you the method. The goal is that after watching the explanation you can solve the next systems problem yourself, not just this one.
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