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Chapter 10: Problem 14

Use the substitution method to find all solutions of each system. $$\left\\{\begin{aligned} 4 x+2 y &=3 \\ 10 x+4 y &=1 \end{aligned}\right.$$

Short Answer

Expert verified
The solution is \(x = -2.5\) and \(y = 6.5\).

Step by step solution

01

Solve for one variable in the first equation

We start with the first equation, which is \(4x + 2y = 3 \). Let's solve for \(y\). Divide the entire equation by 2 to simplify: \(2x + y = 1.5 \). Now, solve for \(y\): \(y = 1.5 - 2x\).
02

Substitute into the second equation

Use the expression for \(y\) from Step 1 and substitute it into the second equation \(10x + 4y = 1\). Substitute \(y = 1.5 - 2x\): \(10x + 4(1.5 - 2x) = 1\). Simplify the equation: \(10x + 6 - 8x = 1\).
03

Simplify and Solve for x

Combine like terms in \(10x + 6 - 8x = 1\) to get \(2x + 6 = 1\). Subtract 6 from both sides to isolate terms with \(x\): \(2x = -5\). Divide by 2: \(x = -2.5\).
04

Substitute x back to find y

Substitute \(x = -2.5\) back into the expression for \(y\) from Step 1: \(y = 1.5 - 2(-2.5)\). Simplify: \(y = 1.5 + 5 = 6.5\).
05

Validate the solution

Substitute \(x = -2.5\) and \(y = 6.5\) back into both original equations to ensure they hold true. For the first equation: \(4(-2.5) + 2(6.5) = -10 + 13 = 3\). For the second equation: \(10(-2.5) + 4(6.5) = -25 + 26 = 1\). Both equations are satisfied, confirming the solution is correct.

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Key Concepts

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

Understanding Systems of Equations
A system of equations consists of multiple equations that share a set of unknowns. In this kind of problem, the goal is to find values for these unknowns that satisfy all equations simultaneously. In our exercise, we are given a system of two linear equations with two variables, \(x\) and \(y\):
  • \(4x + 2y = 3\)
  • \(10x + 4y = 1\)
These equations represent lines when graphed on a coordinate plane. The solution to a system of linear equations typically corresponds to the point at which these lines intersect. An interesting aspect of systems of equations is that they can have:
  • One unique solution (the lines intersect at a single point)
  • No solution (the lines are parallel and never meet)
  • Infinitely many solutions (the lines coincide completely)
By using the substitution method, you can substitute one equation into the other, simplifying the system to find the values of \(x\) and \(y\).
Solving Linear Equations Using Substitution Method
The substitution method is a powerful technique for solving systems of linear equations. It involves substitution to find the solution for all unknowns. In our example, we solve the first equation \(4x + 2y = 3\) for \(y\). This isolates \(y\), which allows us to express it in terms of \(x\):
  • Divide each term by 2 to simplify: \(2x + y = 1.5\)
  • Solve for \(y\): \(y = 1.5 - 2x\)
This new expression for \(y\) is then substituted into the second equation to find \(x\). Doing so transforms the problem into a single-variable equation:
  • For the second equation, substitute: \(10x + 4(1.5 - 2x) = 1\)
  • Simplify and solve for \(x\): after combining like terms, \(2x = -5 \), so \(x = -2.5\)
Using the substitution method effectively simplifies a system of equations by reducing the number of unknowns in a stepwise manner.
Mastering Algebraic Manipulation
Algebraic manipulation involves rearranging equations to isolate one variable or simplify expressions. It's a key skill in solving linear equations. Here鈥檚 how it's applied in our example:
  • Simplification: Initially, we divide the first equation by 2, transforming it into a simpler form: \(2x + y = 1.5\). This makes it easier to express \(y\) in terms of \(x\).

  • Substitution: Once \(y\) is isolated as \(y = 1.5 - 2x\), we substitute this expression into the second equation for the \(y\) variable.

  • Combining like terms: In the modified second equation \(10x + 4(1.5 - 2x) = 1\), distributing and combining like terms helps in simplifying the equation, allowing us to isolate \(x\).

  • Verification: Finally, substituting back and checking that both original equations hold true for \(x = -2.5\) and \(y = 6.5\) confirms the solution is valid.
Algebraic manipulation requires practice, but mastering it empowers you to tackle complex systems of equations with confidence.

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Most popular questions from this chapter

Suppose that the height of an object as a function of time is given by \(f(t)=a t^{2}+b t+c,\) where \(t\) is time in seconds, \(f(t)\) is the height in feet at time \(t,\) and \(a, b,\) and \(c\) are certain constants. If, after \(1,2,\) and 3 seconds, the corresponding heights are \(184 \mathrm{ft}, 136 \mathrm{ft},\) and \(56 \mathrm{ft},\) respectively, find the time at which the object is at ground level (height \(=0 \mathrm{ft}\) ).

The matrices \(A, B, C, D, E, F,\) and \(G\) are defined as follows: $$ A=\left(\begin{array}{rr} 2 & 3 \\ -1 & 4 \end{array}\right) \quad B=\left(\begin{array}{rr} 1 & -1 \\ 3 & 0 \end{array}\right) \quad C=\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) $$ $$\begin{aligned} &D=\left(\begin{array}{rrr} -1 & 2 & 3 \\ 4 & 0 & 5 \end{array}\right) \quad E=\left(\begin{array}{rr} 2 & 1 \\ 8 & -1 \\ 6 & 5 \end{array}\right)\\\ &F=\left(\begin{array}{rr} 5 & -1 \\ -4 & 0 \\ 2 & 3 \end{array}\right) \quad G=\left(\begin{array}{ll} 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{array}\right) \end{aligned}$$ In each exercise, carry out the indicated matrix operations if they are defined. If an operation is not defined, say so. $$A^{2}(=A A)$$

Solve the following system for \(x, y,\) and \(z\) $$ \left\\{\begin{aligned} e^{x}+e^{y}-2 e^{z} &=2 a \\ e^{x}+2 e^{y}-4 e^{z} &=3 a \\ \frac{1}{2} e^{x}-3 e^{y}+e^{z} &=-5 a \end{aligned}\right. $$ (Assume that \(a>0 .) \quad\) Hint: Let \(A=e^{x}, B=e^{y},\) and \(C=e^{z} .\) Solve the resulting linear system for \(A, B,\) and \(C\)

So the input (3,5) yields an output of \(\sqrt{2} .\) We define the do- main for this function just as we did in Chapter 3: The domain is the set of all inputs that yield real-number outputs. For instance, the ordered pair (1,4) is not in the domain of the function we have been discussing, because (as you should check for yourself\() f(1,4)=\sqrt{-1},\) which is not a real number. We can determine the domain of the function in equation ( 1 ) by requiring that the quantity under the radical sign be non negative. Thus we require that \(2 x-y+1 \geq 0\) and, consequently, \(y \leq 2 x+1\) (Check this.) The following figure shows the graph of this inequality; the domain of our function is the set of ordered pairs making up the graph. In Exercises follow a similar procedure and sketch the domain of the given function. (Graph cant copy) $$h(x, y)=\ln (x y)$$

Use Cramers rule to solve those systems for which \(D \neq 0 .\) In cases where \(D=0,\) use Gaussian elimination or matrix methods. $$\left\\{\begin{array}{l} 5 x-3 y-z=16 \\ 2 x+y-3 z=5 \\ 3 x-2 y+2 z=5 \end{array}\right.$$
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