/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Find the intersection in the \(x... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 35

Find the intersection in the \(x y\) -plane of the lines \(y=5 x+3\) and \(y=-2 x+1\)

Short Answer

Expert verified
The intersection point in the $xy$-plane of the lines \(y = 5x + 3\) and \(y = -2x + 1\) is \(\left(-\frac{2}{7}, \frac{11}{7}\right)\).

Step by step solution

01

Set both equations equal to each other

Since we are looking for a point that lies on both lines, the y-coordinates of that point should be equal. Therefore, set the two equations equal to each other to get the following equation: \[5x + 3 = -2x + 1\]
02

Solve for x

The next step is to isolate the x variable. Add 2x to both sides of the equation: \[5x + 2x + 3 = -2x + 2x + 1\] \[7x + 3 = 1\] Now, subtract 3 from both sides of the equation: \[7x + 3 - 3 = 1 - 3\] \[7x = -2\] Finally, to get the value of x, divide both sides of the equation by 7: \[x = -\frac{2}{7}\]
03

Find y by substituting x in either equation

Now that we have found the value of x, we can find y by substituting x in either of the original equations. Let's use the first one: \(y = 5x + 3\) \[y = 5\left(-\frac{2}{7}\right) + 3\] \[y = -\frac{10}{7} + 3\] Now, convert 3 to a fraction with a denominator of 7: \[y = -\frac{10}{7} + \frac{21}{7}\]
04

Calculate the value of y

Add the two fractions together to find the value of y: \[y = -\frac{10}{7} + \frac{21}{7}\] \[y = \frac{11}{7}\]
05

Write the intersection point in coordinate form

The intersection point of the two lines can be written as a coordinate (x, y). So, our final answer is: \[\left(-\frac{2}{7}, \frac{11}{7}\right)\]

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Most popular questions from this chapter

Suppose $$r(x)=\frac{x+1}{x^{2}+3} \quad \text { and } \quad s(x)=\frac{x+2}{x^{2}+5}$$ What is the domain of \(r ?\)

Show that $$ (a+b)^{3}=a^{3}+b^{3} $$ if and only if \(a=0\) or \(b=0\) or \(a=-b\).

Let \(p\) be the polynomial defined by $$p(x)=x^{6}-87 x^{4}-92 x+2$$. (a)Evaluate \(p(-2), p(-1), p(0),\) and \(p(1)\). (b) Explain why the results from part (a) imply that \(p\) has a zero in the interval (-2,-1) and \(p\) has a zero in the interval (0,1) . (c) Show that \(p\) has at least four zeros in the interval [-10,10] . [Hint: We already know from part ( \(b\) ) that \(p\) has at least two zeros is the interval [-10,10] . You can show the existence of other zeros by finding integers \(n\) such that one of the numbers \(p(n)\), \(p(n+1)\) is positive and the other is negative.]

Let \(p\) be the polynomial defined by $$p(x)=x^{6}-87 x^{4}-92 x+2$$. (a)Use a computer or calculator to sketch a graph of \(p\) on the interval [-5,5] . (b) Is \(p(x)\) positive or negative for \(x\) near \(\infty ?\) (c) Is \(p(x)\) positive or negative for \(x\) near \(-\infty ?\) (d) Explain why the graph from part (a) does not accurately show the behavior of \(p(x)\) for large values of \(x\).

Give an example of a polynomial \(p\) of degree 8 such that \(p(2)=3\) and \(p(x) \geq 3\) for all real numbers \(x\).
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