Chapter 4: Problem 50
Solve each system by substitution. If a system has no solution or infinitely many solutions, so state. $$ \left\\{\begin{array}{l} {2 b-a=-1} \\ {3 a+10 b=-1} \end{array}\right. $$
Short Answer
Expert verified
The solution is \(a = \frac{1}{2}\) and \(b = -\frac{1}{4}\).
Step by step solution
01
Express one variable in terms of the other
From the first equation, \(2b - a = -1\), express \(a\) in terms of \(b\). By rearranging the equation, we have \(a = 2b + 1\).
02
Substitute into the second equation
Substitute \(a = 2b + 1\) into the second equation \(3a + 10b = -1\). The substitution gives us \(3(2b + 1) + 10b = -1\).
03
Simplify substituted equation
Distribute the 3 in the substituted equation: \(6b + 3 + 10b = -1\). This simplifies to \(16b + 3 = -1\).
04
Solve for the variable
Isolate \(b\) by subtracting 3 from both sides: \(16b = -4\). Divide by 16: \(b = -\frac{1}{4}\).
05
Substitute back to find the other variable
Substitute \(b = -\frac{1}{4}\) back into the equation \(a = 2b + 1\): \(a = 2(-\frac{1}{4}) + 1 = -\frac{1}{2} + 1 = \frac{1}{2}\).
06
Verify the solution
Substitute \(a = \frac{1}{2}\) and \(b = -\frac{1}{4}\) into both original equations to ensure they are satisfied. Substituting into the first equation: \(2(-\frac{1}{4}) - \frac{1}{2} = -\frac{1}{2} - \frac{1}{2} = -1\), which is correct. Substituting into the second equation: \(3(\frac{1}{2}) + 10(-\frac{1}{4}) = \frac{3}{2} - \frac{10}{4} = -1\), which is also correct.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Systems of Equations
A system of equations is a collection of two or more equations with a common set of variables. In this exercise, we are dealing with two equations with two unknowns, namely \(a\) and \(b\).
- The goal is to find values for \(a\) and \(b\) that make both equations true at the same time.
- These types of problems are common in algebra and are usually solved by methods such as substitution, elimination, or graphing.
- One unique solution: both equations intersect at a single point.
- No solution: the lines are parallel and do not intersect.
- Infinite solutions: the lines overlap completely.
Linear Equations
Linear equations are algebraic expressions representing straight lines when graphed. The general form of a linear equation in two variables is \(ax + by = c\). In this exercise, both equations are linear, meaning each represents a straight line.
- In the given system, the first equation is \(2b - a = -1\) and the second is \(3a + 10b = -1\).
- The coefficients (numbers multiplying the variables) define the slope and position of each line on a graph.
Solution Verification
Verification is an essential step in solving systems of equations, ensuring that the found solutions satisfy all original equations.
- It's crucial because errors can occur during algebraic manipulation, and verifying helps catch these mistakes.