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Solve the system by the method of substitution. $$\left\\{\begin{aligned} 6 x-3 y-4 &=0 \\ x+2 y-4 &=0 \end{aligned}\right.$$

Short Answer

Expert verified
The solution to the system of equations is \(x = 4 / 3\) and \(y = 4 / 3\).

Step by step solution

01

Isolate a variable

Choose one of the equations and isolate one variable. Let's choose the second equation \(x + 2y - 4 = 0\), isolate \(x\): \(x = 4 - 2y\).
02

Substitute into the other equation

Now, substitute \(x = 4 - 2y\) into the first given equation \(6x - 3y - 4 = 0\) resulting in \(6(4 - 2y) - 3y - 4 = 0\). Simplifying this equation gives \(24 - 12y - 3y - 4 = 0\), which further simplifies to \(-15y + 20 = 0\).
03

Solve for the isolated variable

Solve \(-15y + 20 = 0\) for \(y\) gives \(y = 20 / 15 = 4 / 3\).
04

Substitute back into rearranged equation

Now we know \(y = 4 / 3\), substitute this into the rearranged equation from step 1: \(x = 4 - 2*(4 / 3)\). Solving for \(x\) gives \(x = 4 / 3\).

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Key Concepts

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

Method of Substitution
The method of substitution is a fundamental algebraic technique used to solve systems of linear equations. It involves replacing one variable with an equivalent expression obtained from another equation within the system. This technique is particularly useful when one equation can be easily solved for one variable in terms of the others.

For instance, with our given system of equations, the second equation is chosen to isolate variable 'x'. Once 'x' is expressed in terms of 'y', this expression is substituted into the first equation in place of 'x'. This reduces the system from two equations in two variables to a single equation in one variable which is then much easier to solve. Using substitution avoids the need to manipulate both equations simultaneously, thus simplifying the problem-solving process.
Solving Systems of Linear Equations
Solving systems of linear equations is essential for finding the values of unknowns that satisfy all equations in the system simultaneously. Put another way, we are looking for the point(s) of intersection between lines, planes, or hyperplanes represented by these equations in n-dimensional space.

To solve such a system, we often leverage methods like graphing, substitution, elimination, and matrix operations. The substitution method, in particular, is advantageous when the system includes an equation that can be conveniently solved for one variable. Through step-by-step replacement and simplification, the system's complexity is broken down until all variable values are determined, yielding a solution that satisfies all the original conditions.
Isolation of Variables
Isolation of variables involves manipulating an equation to have the subject variable on one side of the equation, set apart from the other variables or constants. This is especially helpful when preparing to use the substitution method for solving systems of equations.

In our exercise, the first step is critical: choosing the most suitable equation to isolate a variable. By isolating 'x' in the second equation, we set up a straightforward pathway to substitute this expression into other equations of the system. This technique enables a clear and sequential path to solution, as it minimizes the potential for algebraic error and clarifies the problem-solving process. The goal is to make the equation look like 'variable = expression', ensuring a smooth substitution.
Algebraic Substitution
Algebraic substitution is the act of replacing variables with their equivalent values or expressions. After isolating a variable in one equation, we substitute that expression into another equation—exactly what happens within the substitution method.

Returning to our example, after isolating 'x', we substitute the expression for 'x' into the first equation. This reshapes the equation into a single variable format, in this case, 'y' only, which can then be solved directly. Algebraic substitution is not only used in solving systems of equations but also in simplifying expressions, integrating functions, and many other contexts within mathematics. It's a versatile tool that reduces problems to a more manageable size by methodically eliminating variables one by one.

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Most popular questions from this chapter

Optimal Profit A manufacturer produces two models of elliptical cross-training exercise machines. The times for assembling, finishing, and packaging model \(\mathrm{A}\) are 3 hours, 3 hours, and \(0.8\) hour, respectively. The times for model B are 4 hours, \(2.5\) hours, and \(0.4\) hour. The total times available for assembling, finishing, and packaging are 6000 hours, 4200 hours, and 950 hours, respectively. The profits per unit are \(\$ 300\) for model \(A\) and $$\$ 375$$ for model \(B\). What is the optimal production level for each model? What is the optimal profit?

MAKE A DECISION: DIET SUPPLEMENT A dietitian designs a special diet supplement using two different foods. Each ounce of food \(\mathrm{X}\) contains 12 units of calcium, 10 units of iron, and 20 units of vitamin \(\mathrm{B}\). Each ounce of food \(\mathrm{Y}\) contains 15 units of calcium, 20 units of iron, and 12 units of vitamin B. The minimum daily requirements for the diet are 300 units of calcium, 280 units of iron, and 300 units of vitamin \(\mathrm{B}\). (a) Find a system of inequalities describing the different amounts of food \(\mathrm{X}\) and food \(\mathrm{Y}\) that the dietitian can use in the diet. (b) Sketch the graph of the system. (c) A nutritionist normally gives a patient 10 ounces of food \(\mathrm{X}\) and 12 ounces of food \(\mathrm{Y}\) per day. Supplies of food \(\mathrm{Y}\) are running low. What other combinations of foods \(\mathrm{X}\) and \(\mathrm{Y}\) can be given to the patient to meet the minimum daily requirements?

Investment You plan to invest up to $$\$ 30,000$$ in two different interest- bearing accounts. Each account is to contain at least $$\$ 6000$$. Moreover, one account should have at least twice the amount that is in the other account. (a) Find a system of inequalities that describes the amounts that you can invest in each account, and (b) sketch the graph of the system.

Sketch the region determined by the constraints. Then find the minimum anc maximum values of the objective function and where they occur, subject to the indicated constraints. Objective function: $$ z=x $$ Constraints: $$ \begin{array}{r} x \geq 0 \\ y \geq 0 \\ 2 x+3 y \leq 60 \\ 2 x+y \leq 28 \\ 4 x+y \leq 48 \end{array} $$

Sketch the graph of the inequality. $$5 x+3 y \geq-15$$

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