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Solve the system by either the substitution or th…

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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 53 Problem 54 Problem 55 Problem 56 Problem 57 Problem 58 Problem 59 Problem 60 Problem 61 Problem 62 Problem 63 Problem 64 Problem 65 Problem 66 Problem 67 Problem 68 Problem 69 Problem 70 Problem 71 Problem 72 Problem 73 Problem 74 Problem 75 Problem 76 Problem 77 Problem 78 Problem 79 Problem 80 Problem 81 Problem 82 Problem 83 Problem 84 Problem 85 Problem 86 Problem 87 Problem 88 Problem 89 Problem 90 Problem 91 Problem 92 Problem 93 Problem 94 Problem 95 Problem 96 Problem 97 Problem 98 Problem 99 Problem 100 Problem 101 Problem 102 Problem 103 Problem 104

Problem 50 Hard Difficulty

Solve the system by either the substitution or the elimination method.
$$
\left\{\begin{array}{l}
{9 x+4 y=31} \\
{y-5=6 x}
\end{array}\right.
$$

Answer

$x=1 / 3$
$y=7$

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Elementary and Intermediate Algebra

Chapter 4

Systems of Linear Equations and Inequalities

Section 3

Solving Systems of Equations by Elimination (Addition)

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Video Transcript

So for this problem, the way it's set up, I'm going to go ahead and use substitution. I'm gonna start by eliminating or isolating why in the second equation and I'm gonna add five to both sides and I get why equals six X plus five. I'll call that equation too, and we'll call this appear equation one. So now what we want to do is take equation one, and we want to substitute six X plus five in for the why. So Equation one is now nine x plus four times the quantity six x plus five equals 31. And we're gonna simplify that first by distributing the four so nine X plus 24 x plus 20 equals 31. And now we combine the light terms nine X plus 24 x is 33 x so 33 X plus 20 equals 31. We can subtract the 20 from both sides and we have 33 x equals 11 and then divide both sides by three and weak 33 we get X equals 1/3 now to find the value of why we can use equation too, and substitute our X value in and we have y equals six times, 1/3 plus five. So that works out to be y equals two plus five. So Y equals seven. That's a good idea to do a little check, and I'm gonna check using the first equation and plugging in both x and y values. So nine times 1/3 plus four times seven equals 2131 Sometimes 1/3 is 34 times seven is 28 3 times 28 3 plus 28 does equal 31. So we're good there. That check worked. So now we write our answer as an ordered pair x comma y 1/3 comma seven.

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Alan S. Tussy, R. David Gustafson

Elementary and Intermediate Algebra

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