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Use the method of substitution to solve the system. $$\left\\{\begin{aligned} x+2 y &=-1 \\ 2 x-3 y &=12 \end{aligned}\right.$$

Short Answer

Expert verified
The solution is \( x = 3 \) and \( y = -2 \).

Step by step solution

01

Solve the First Equation for x

The first equation is \( x + 2y = -1 \). Solve for \( x \) in terms of \( y \). This gives \( x = -1 - 2y \). This expression for \( x \) will be used in the next step.
02

Substitute Expression into Second Equation

Substitute \( x = -1 - 2y \) into the second equation \( 2x - 3y = 12 \). This gives \( 2(-1 - 2y) - 3y = 12 \).
03

Simplify and Solve for y

Simplify the equation: \( -2 - 4y - 3y = 12 \), which simplifies to \( -2 - 7y = 12 \). Add 2 to both sides: \( -7y = 14 \). Divide by \( -7 \): \( y = -2 \).
04

Substitute y back to Find x

Substitute \( y = -2 \) into the expression for \( x \): \( x = -1 - 2(-2) \). Simplify: \( x = -1 + 4 = 3 \).
05

Verify the Solution

Substitute \( x = 3 \) and \( y = -2 \) back into the original equations to verify. For the first equation: \( 3 + 2(-2) = -1 \), which holds true. For the second equation: \( 2(3) - 3(-2) = 12 \), which is also true. The solution is verified.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

System of Linear Equations
A system of linear equations consists of two or more linear equations with the same set of variables. In our case, we have a system with two linear equations:
  • \( x + 2y = -1 \)
  • \( 2x - 3y = 12 \)
The solution to a system of linear equations is the set of values for the variables that satisfy all equations in the system simultaneously. Generally, these systems can have:
  • One unique solution, if the equations intersect at a single point.
  • No solution, if the equations represent parallel lines.
  • Infinitely many solutions, if the equations are identical, representing the same line.
In our exercise, the goal is to find the unique values of \( x \) and \( y \) using the substitution method, which we'll explain in detail in the following sections.
Solving Equations
Solving an equation means finding the value(s) for the variable(s) that make the equation true. For linear equations, this typically involves manipulating the equation using algebraic operations to isolate one or more variables.
In the substitution method, the process begins with solving one of the equations for one variable in terms of another. This involves simple algebraic steps like:
  • Adding or subtracting terms from both sides of the equation.
  • Multiplying or dividing both sides of the equation by a number.
In our example, the first step was to solve the equation \( x + 2y = -1 \) for \( x \), giving us \( x = -1 - 2y \). This expression for \( x \) allows us to substitute into the other equation, aiding in solving for the other variable.
Substitution Technique
The substitution technique is a powerful tool for solving systems of linear equations. It involves substituting an expression derived from one equation into another. Here’s the step-by-step breakdown:
First, choose one of the equations and solve it for one variable. This can be the easier equation out of the two. In our problem, we opted for \( x = -1 - 2y \) from the first equation.
Next, substitute the expression for this variable into the other equation. Replace \( x \) in \( 2x - 3y = 12 \) with \( -1 - 2y \), resulting in a single equation in terms of \( y \).
Now, solve this new equation for \( y \). This involved simplifying \( 2(-1 - 2y) - 3y = 12 \) to \( y = -2 \).
Finally, use the value of \( y \) to find \( x \) by substituting \( y = -2 \) back into the earlier expression \( x = -1 - 2y \). This gives us \( x = 3 \).
Verification is key: Plugging these values back into the original equations confirmed they are correct. Both equations are satisfied with \( x = 3 \) and \( y = -2 \), showing the efficiency and accuracy of the substitution technique.

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Most popular questions from this chapter

Let \(I=I_{2}\) be the identity matrix of order 2, and let \(f(x)=|\boldsymbol{A}-\boldsymbol{x} \boldsymbol{I}| .\) Find (a) the polynomial \(f(\boldsymbol{x})\) and (b) the zeros of \(f(x)\). (In the study of matrices, \(f(x)\) is the characteristic polynomial of \(A,\) and the zeros of \(f(x)\) are the characteristic values (eigenvalues) of \(A .\) ) $$A=\left[\begin{array}{ll} 3 & 1 \\ 2 & 2 \end{array}\right]$$

There are three chains, weighing 450 \(610,\) and 950 ounces, each consisting of links of three different sizes. Each chain has 10 small links. The chains also have \(20,30,\) and 40 medium links and \(30,40,\) and 70 large links, respectively. Find the weights of the small, medium, and large links.

Use the method of substitution to solve the system. $$\left\\{\begin{array}{l} 6 x^{3}-y^{3}=1 \\ 3 x^{3}+4 y^{3}=5 \end{array}\right.$$

Let \(I=I_{2}\) be the identity matrix of order 2, and let \(f(x)=|\boldsymbol{A}-\boldsymbol{x} \boldsymbol{I}| .\) Find (a) the polynomial \(f(\boldsymbol{x})\) and (b) the zeros of \(f(x)\). (In the study of matrices, \(f(x)\) is the characteristic polynomial of \(A,\) and the zeros of \(f(x)\) are the characteristic values (eigenvalues) of \(A .\) ) $$A=\left[\begin{array}{rr} -3 & -2 \\ 2 & 2 \end{array}\right]$$

Let $$A=\left[\begin{array}{rrr}3 & -3 & 7 \\\2 & 6 & -2 \\\4 & 2 & 5\end{array}\right] \text { and }B=\left[\begin{array}{rrr}-9 & 5 & -8 \\\3 & -7 & 1 \\\\-1 & 2 & 6\end{array}\right]$$ Evaluate the matrix expression. $$A^{2}-5 B$$

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