Can't solve system of linear equations (that need simplification first) Announcing the arrival...
Why did Europeans not widely domesticate foxes?
Putting Ant-Man on house arrest
How long can a nation maintain a technological edge over the rest of the world?
What does the black goddess statue do and what is it?
How to keep bees out of canned beverages?
Why isn't everyone flabbergasted about Bran's "gift"?
Raising a bilingual kid. When should we introduce the majority language?
When speaking, how do you change your mind mid-sentence?
When I export an AI 300x60 art board it saves with bigger dimensions
What were wait-states, and why was it only an issue for PCs?
What helicopter has the most rotor blades?
Test if all elements of a Foldable are the same
Marquee sign letters
What do you call an IPA symbol that lacks a name (e.g. ɲ)?
How to begin with a paragraph in latex
How to translate "red flag" into Spanish?
The 'gros' functor from schemes into (strictly) locally ringed topoi
France's Public Holidays' Puzzle
Why doesn't the university give past final exams' answers?
What is the purpose of the side handle on a hand ("eggbeater") drill?
What *exactly* is electrical current, voltage, and resistance?
Suing a Police Officer Instead of the Police Department
When does Bran Stark remember Jamie pushing him?
Is there a possibility to generate a list dynamically in Latex?
Can't solve system of linear equations (that need simplification first)
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30 pm US/Eastern)System of linear equations in 8 variablesChecking if systems of linear equations are equivalentLinear system of equations of three variables confusion about determinismSolve nonlinear system of equationsSystem of two quadratic equationsFinding a unified ratio from two separate ratiosHow do we get from $ln A=ln P+rn$ to $A=Pe^{rn}$ and similar logarithmic equations?Dividing higher-order algebraic expressionsSimplifying a polynomialSimplyfying $(x^2-y^2)^frac{1}{2}times(x-y)^frac{3}{2}times(x+y)^frac{-1}{2}$ algebraic experssion
$begingroup$
I'm self-studying from Stroud & Booth's amazing "Engineering Mathematics", and am stuck on a problem at the end of the "Linear Equations" chapter.
I've a system of two linear equations:
$$frac{3x+2}{4} - frac{x+2y}{2} = frac{x-3}{12}$$
$$frac{2y+1}{5} + frac{x-3y}{4} = frac{3x+1}{10}$$
So, these two first need to be simplified. I assume that the LSM for the first one (for 2, 4 and 12) is 12, so we have:
$$12frac{3x+2}{4} - 12frac{x+2y}{2} = 12frac{x-3}{12}$$
Simplifying further, we have:
$$3(3x+2) - 6(x+2y) = (x-3)$$
$$9x + 6 - 6x - 12 y = x - 3$$
$$9x -6x -x - 12y = -3 +6$$
Finally, we get our first simplified linear equation:
$$2x - 12y = 3$$
Now, onto the second one. The LSM of 5, 4 and 10 is 20, so we have:
$$20frac{2y+1}{5} + 20frac{x-3y}{4} = 20frac{3x+1}{10}$$
Simplifying further, we have:
$$4(2y+1) + 5(x-3y) = 2(3x + 1)$$
$$8y + 4 + 5x - 15y = 6x + 2$$
$$5x - 6x + 8y - 15y = 2 - 4$$
We get our second simplified linear equation:
$$-x -7y = 2$$
Now we can solve our system of linear equations:
$$2x - 12y = 3$$
$$-x -7y = 4$$
Multiplying the second one by 2:
$$2x - 12y = 3$$
$$-2x -14y = 4$$
Now, we add the two equations, and get:
$$-26y = 7$$
Solving for $y$, we get:
$$y = -frac{7}{26}$$
which I'm fairly certain is not a correct answer.
Can anyone see where I'm going wrong here?
algebra-precalculus self-learning
asked 11 hours ago
neuronneuron
25817
$endgroup$
add a comment |
$begingroup$
I'm self-studying from Stroud & Booth's amazing "Engineering Mathematics", and am stuck on a problem at the end of the "Linear Equations" chapter.
I've a system of two linear equations:
$$frac{3x+2}{4} - frac{x+2y}{2} = frac{x-3}{12}$$
$$frac{2y+1}{5} + frac{x-3y}{4} = frac{3x+1}{10}$$
So, these two first need to be simplified. I assume that the LSM for the first one (for 2, 4 and 12) is 12, so we have:
$$12frac{3x+2}{4} - 12frac{x+2y}{2} = 12frac{x-3}{12}$$
Simplifying further, we have:
$$3(3x+2) - 6(x+2y) = (x-3)$$
$$9x + 6 - 6x - 12 y = x - 3$$
$$9x -6x -x - 12y = -3 +6$$
Finally, we get our first simplified linear equation:
$$2x - 12y = 3$$
Now, onto the second one. The LSM of 5, 4 and 10 is 20, so we have:
$$20frac{2y+1}{5} + 20frac{x-3y}{4} = 20frac{3x+1}{10}$$
Simplifying further, we have:
$$4(2y+1) + 5(x-3y) = 2(3x + 1)$$
$$8y + 4 + 5x - 15y = 6x + 2$$
$$5x - 6x + 8y - 15y = 2 - 4$$
We get our second simplified linear equation:
$$-x -7y = 2$$
Now we can solve our system of linear equations:
$$2x - 12y = 3$$
$$-x -7y = 4$$
Multiplying the second one by 2:
$$2x - 12y = 3$$
$$-2x -14y = 4$$
Now, we add the two equations, and get:
$$-26y = 7$$
Solving for $y$, we get:
$$y = -frac{7}{26}$$
which I'm fairly certain is not a correct answer.
Can anyone see where I'm going wrong here?
algebra-precalculus self-learning
asked 11 hours ago
neuronneuron
25817
$endgroup$
$begingroup$
When you take the $+6$ from the left hand side to the right hand side of the equal sign in the first equation you must switch its sign and it wil become $-6$. And in the second equation $2-4=-2$ and not $2$. So you'll get that $y=frac{1}{2}$ at the end
$endgroup$
– Fareed AF
11 hours ago
add a comment |
$begingroup$
I'm self-studying from Stroud & Booth's amazing "Engineering Mathematics", and am stuck on a problem at the end of the "Linear Equations" chapter.
I've a system of two linear equations:
$$frac{3x+2}{4} - frac{x+2y}{2} = frac{x-3}{12}$$
$$frac{2y+1}{5} + frac{x-3y}{4} = frac{3x+1}{10}$$
So, these two first need to be simplified. I assume that the LSM for the first one (for 2, 4 and 12) is 12, so we have:
$$12frac{3x+2}{4} - 12frac{x+2y}{2} = 12frac{x-3}{12}$$
Simplifying further, we have:
$$3(3x+2) - 6(x+2y) = (x-3)$$
$$9x + 6 - 6x - 12 y = x - 3$$
$$9x -6x -x - 12y = -3 +6$$
Finally, we get our first simplified linear equation:
$$2x - 12y = 3$$
Now, onto the second one. The LSM of 5, 4 and 10 is 20, so we have:
$$20frac{2y+1}{5} + 20frac{x-3y}{4} = 20frac{3x+1}{10}$$
Simplifying further, we have:
$$4(2y+1) + 5(x-3y) = 2(3x + 1)$$
$$8y + 4 + 5x - 15y = 6x + 2$$
$$5x - 6x + 8y - 15y = 2 - 4$$
We get our second simplified linear equation:
$$-x -7y = 2$$
Now we can solve our system of linear equations:
$$2x - 12y = 3$$
$$-x -7y = 4$$
Multiplying the second one by 2:
$$2x - 12y = 3$$
$$-2x -14y = 4$$
Now, we add the two equations, and get:
$$-26y = 7$$
Solving for $y$, we get:
$$y = -frac{7}{26}$$
which I'm fairly certain is not a correct answer.
Can anyone see where I'm going wrong here?
algebra-precalculus self-learning
asked 11 hours ago
neuronneuron
25817
$endgroup$
I'm self-studying from Stroud & Booth's amazing "Engineering Mathematics", and am stuck on a problem at the end of the "Linear Equations" chapter.
I've a system of two linear equations:
$$frac{3x+2}{4} - frac{x+2y}{2} = frac{x-3}{12}$$
$$frac{2y+1}{5} + frac{x-3y}{4} = frac{3x+1}{10}$$
So, these two first need to be simplified. I assume that the LSM for the first one (for 2, 4 and 12) is 12, so we have:
$$12frac{3x+2}{4} - 12frac{x+2y}{2} = 12frac{x-3}{12}$$
Simplifying further, we have:
$$3(3x+2) - 6(x+2y) = (x-3)$$
$$9x + 6 - 6x - 12 y = x - 3$$
$$9x -6x -x - 12y = -3 +6$$
Finally, we get our first simplified linear equation:
$$2x - 12y = 3$$
Now, onto the second one. The LSM of 5, 4 and 10 is 20, so we have:
$$20frac{2y+1}{5} + 20frac{x-3y}{4} = 20frac{3x+1}{10}$$
Simplifying further, we have:
$$4(2y+1) + 5(x-3y) = 2(3x + 1)$$
$$8y + 4 + 5x - 15y = 6x + 2$$
$$5x - 6x + 8y - 15y = 2 - 4$$
We get our second simplified linear equation:
$$-x -7y = 2$$
Now we can solve our system of linear equations:
$$2x - 12y = 3$$
$$-x -7y = 4$$
Multiplying the second one by 2:
$$2x - 12y = 3$$
$$-2x -14y = 4$$
Now, we add the two equations, and get:
$$-26y = 7$$
Solving for $y$, we get:
$$y = -frac{7}{26}$$
which I'm fairly certain is not a correct answer.
Can anyone see where I'm going wrong here?
algebra-precalculus self-learning
algebra-precalculus self-learning
asked 11 hours ago
neuronneuron
25817
asked 11 hours ago
neuronneuron
25817
edited 6 hours ago
neuron
asked 11 hours ago
neuronneuron
25817
asked 11 hours ago
neuronneuron
25817
asked 11 hours ago
neuronneuron
25817
25817
$begingroup$
When you take the $+6$ from the left hand side to the right hand side of the equal sign in the first equation you must switch its sign and it wil become $-6$. And in the second equation $2-4=-2$ and not $2$. So you'll get that $y=frac{1}{2}$ at the end
$endgroup$
– Fareed AF
11 hours ago
add a comment |
$begingroup$
When you take the $+6$ from the left hand side to the right hand side of the equal sign in the first equation you must switch its sign and it wil become $-6$. And in the second equation $2-4=-2$ and not $2$. So you'll get that $y=frac{1}{2}$ at the end
$endgroup$
– Fareed AF
11 hours ago
$begingroup$
When you take the $+6$ from the left hand side to the right hand side of the equal sign in the first equation you must switch its sign and it wil become $-6$. And in the second equation $2-4=-2$ and not $2$. So you'll get that $y=frac{1}{2}$ at the end
$endgroup$
– Fareed AF
11 hours ago
$begingroup$
When you take the $+6$ from the left hand side to the right hand side of the equal sign in the first equation you must switch its sign and it wil become $-6$. And in the second equation $2-4=-2$ and not $2$. So you'll get that $y=frac{1}{2}$ at the end
$endgroup$
– Fareed AF
11 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The first mistake I see is here:
$$9x + 6 - 6x - 12 y = x - 3$$ $$9x -6x -x - 12y = -3 +6$$
The second line should read
$$9x-6x-x-12y=-3-6$$
you forgot to invert the sign when moving $6$ to the other side of the $=$ sign.
You were equally sloppy with the second equation, when you did this:
$$5x - 6x + 8y - 15y = 2 - 4$$
We get our second simplified linear equation: $$-x -7y = 2$$
In fact, $2-4$ is not equal to $2$.
answered 11 hours ago
5xum5xum
93.1k395162
$endgroup$
add a comment |
$begingroup$
The first equation simplifies to $$2x-12y=-9$$ and the second one to $$-7y-x=-2$$
answered 11 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
$endgroup$
add a comment |
$begingroup$
Finally, we get our first simplified linear equation:
This step is wrong:
$2x-12y=-9$ this should be your first equation.
answered 11 hours ago
HS SinghHS Singh
112
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3198159%2fcant-solve-system-of-linear-equations-that-need-simplification-first%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The first mistake I see is here:
$$9x + 6 - 6x - 12 y = x - 3$$ $$9x -6x -x - 12y = -3 +6$$
The second line should read
$$9x-6x-x-12y=-3-6$$
you forgot to invert the sign when moving $6$ to the other side of the $=$ sign.
You were equally sloppy with the second equation, when you did this:
$$5x - 6x + 8y - 15y = 2 - 4$$
We get our second simplified linear equation: $$-x -7y = 2$$
In fact, $2-4$ is not equal to $2$.
answered 11 hours ago
5xum5xum
93.1k395162
$endgroup$
add a comment |
$begingroup$
The first mistake I see is here:
$$9x + 6 - 6x - 12 y = x - 3$$ $$9x -6x -x - 12y = -3 +6$$
The second line should read
$$9x-6x-x-12y=-3-6$$
you forgot to invert the sign when moving $6$ to the other side of the $=$ sign.
You were equally sloppy with the second equation, when you did this:
$$5x - 6x + 8y - 15y = 2 - 4$$
We get our second simplified linear equation: $$-x -7y = 2$$
In fact, $2-4$ is not equal to $2$.
answered 11 hours ago
5xum5xum
93.1k395162
$endgroup$
add a comment |
$begingroup$
The first mistake I see is here:
$$9x + 6 - 6x - 12 y = x - 3$$ $$9x -6x -x - 12y = -3 +6$$
The second line should read
$$9x-6x-x-12y=-3-6$$
you forgot to invert the sign when moving $6$ to the other side of the $=$ sign.
You were equally sloppy with the second equation, when you did this:
$$5x - 6x + 8y - 15y = 2 - 4$$
We get our second simplified linear equation: $$-x -7y = 2$$
In fact, $2-4$ is not equal to $2$.
answered 11 hours ago
5xum5xum
93.1k395162
$endgroup$
The first mistake I see is here:
$$9x + 6 - 6x - 12 y = x - 3$$ $$9x -6x -x - 12y = -3 +6$$
The second line should read
$$9x-6x-x-12y=-3-6$$
you forgot to invert the sign when moving $6$ to the other side of the $=$ sign.
You were equally sloppy with the second equation, when you did this:
$$5x - 6x + 8y - 15y = 2 - 4$$
We get our second simplified linear equation: $$-x -7y = 2$$
In fact, $2-4$ is not equal to $2$.
answered 11 hours ago
5xum5xum
93.1k395162
answered 11 hours ago
5xum5xum
93.1k395162
answered 11 hours ago
5xum5xum
93.1k395162
answered 11 hours ago
5xum5xum
93.1k395162
93.1k395162
add a comment |
add a comment |
$begingroup$
The first equation simplifies to $$2x-12y=-9$$ and the second one to $$-7y-x=-2$$
answered 11 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
$endgroup$
add a comment |
$begingroup$
The first equation simplifies to $$2x-12y=-9$$ and the second one to $$-7y-x=-2$$
answered 11 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
$endgroup$
add a comment |
$begingroup$
The first equation simplifies to $$2x-12y=-9$$ and the second one to $$-7y-x=-2$$
answered 11 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
$endgroup$
The first equation simplifies to $$2x-12y=-9$$ and the second one to $$-7y-x=-2$$
answered 11 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
answered 11 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
answered 11 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
answered 11 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.6k42867
79.6k42867
add a comment |
add a comment |
$begingroup$
Finally, we get our first simplified linear equation:
This step is wrong:
$2x-12y=-9$ this should be your first equation.
answered 11 hours ago
HS SinghHS Singh
112
$endgroup$
add a comment |
$begingroup$
Finally, we get our first simplified linear equation:
This step is wrong:
$2x-12y=-9$ this should be your first equation.
answered 11 hours ago
HS SinghHS Singh
112
$endgroup$
add a comment |
$begingroup$
Finally, we get our first simplified linear equation:
This step is wrong:
$2x-12y=-9$ this should be your first equation.
answered 11 hours ago
HS SinghHS Singh
112
$endgroup$
Finally, we get our first simplified linear equation:
This step is wrong:
$2x-12y=-9$ this should be your first equation.
answered 11 hours ago
HS SinghHS Singh
112
answered 11 hours ago
HS SinghHS Singh
112
answered 11 hours ago
HS SinghHS Singh
112
answered 11 hours ago
HS SinghHS Singh
112
112
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3198159%2fcant-solve-system-of-linear-equations-that-need-simplification-first%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
When you take the $+6$ from the left hand side to the right hand side of the equal sign in the first equation you must switch its sign and it wil become $-6$. And in the second equation $2-4=-2$ and not $2$. So you'll get that $y=frac{1}{2}$ at the end
$endgroup$
– Fareed AF
11 hours ago