Master Algebra 1 Word Problems with our Comprehensive Practice

A question defines
${\displaystyle f\left(x\right)=(ax+b)e^x}$
and states that it satisfies the equation
${\displaystyle f\left(x\right)={\int }_0^x{e}^{x-y}{f}^{\prime }\left(y\right)dy-(x^2-x+1)e^x}$
You're required to find a and b.

A person is making a 5-gallon punch for a party that will consist of two different types of juices: apple and orange juice. They use at least 1 gallon of each type but want to use more apple than orange juice.
x represents the gallons of apple juice in the punch and y represents the gallons of orange juice in the punch. Create a system of equations and/or inequalities modeling the scenario.

A group of 7 students go out for lunch. If two students have hamburgers and five students have hot dogs, the bill will be 18.75 dollars. If five students have hamburgers and two students have hot dogs, the bill will be 23.25 dollars. What is the price of a hamburger?

Complete the equation of the line through (-10,3) (-8,-8)

Does ${\displaystyle x^3-x^2}$ equal ${\displaystyle x^2}$? How can I prove so, mathematically?

Let ${\displaystyle A\in M_{m\times n}\left(\mathbb{Q}\right)}$ and ${\displaystyle b\in {\mathbb{Q}}^m}$. Suppose that the system of linear equations Ax=b has a solution in ${\displaystyle {\mathbb{R}}^n}$. Does it necessarily have a solution in ${\displaystyle {\mathbb{Q}}^n}$?
and I thought I'd give an interesting, possibly wrong, approach to solving it. I'm not sure if such things can be done, if not maybe you can help me refine.
I considered the form of the equality as
${\displaystyle A^{\left(1\right)}{x}_1+\cdots +A^{\left(n\right)}{x}_n=b}$
where ${\displaystyle A^{\left(i\right)}}$ is a column vector of A. I then noticed that for ${\displaystyle x_i\in \mathbb{R}\mathrm{\setminus }\mathbb{Q}=\mathbb{T}}$ then, and this is where I think I'm doing something forbidden, each x has the represenation
${\displaystyle x_1=k_{11}{\tau }_1+\cdots +k_{1p}{\tau }_p}$
${\displaystyle x_2=k_{21}{\tau }_1+\cdots +k_{2p}{\tau }_p}$
${\displaystyle ⋮}$
${\displaystyle x_n=k_{n1}{\tau }_1+\cdots +k_{np}{\tau }_p}$,
where ${\displaystyle {\tau }_i}$ is a distinct irrational number, ${\displaystyle k_{ij}\in \mathbb{R}}$, and p is the number of such distinct irrational numbers. I wound this out, but there may be a discrepancy with p and m. I feel this method can lead me to the answer, but I'm not sure where to go from here.
I end up getting something like this, I believe, after substitution:
${\displaystyle A(k^{\left(1\right)}{\tau }_1+\cdots +k^{\left(p\right)}{\tau }_p)=b}$
Here, ${\displaystyle k^{\left(i\right)}}$ is the vector
$((k_{1i}),(\dots ),(k_{ni}))$.
I think there is no discrepancy with p and m because ${\displaystyle A\in M_{m\times n}\left(\mathbb{Q}\right),K\in M_{n\times p}\left(\mathbb{R}\right)}$, and ${\displaystyle \tau \in M_{p\times 1}\left(\mathbb{T}\right)}$, so
${\displaystyle (m\times n)\cdot (n\times p)\cdot (p\times 1)=m\times 1}$.

The problem is:
The points (2,-1,-2), (1,3,12), and (4,2,3) lie on a unique plane. Where does this plane cross the z-axis?
I can easily solve this problem by calculus and cross product: the equation of a plane is
${\displaystyle -22x+33x-11z=-55.}$
Hence, this plane crosses the z-axis at z=5. However, the problem requires to be sold by the system of linear equations transformed into matrix. Even vectors have not been introduces yet. Just a matrix in echelon form and a back substitution. Hence, when I am writing three equations:
${\displaystyle 2a-b-2c=d}$
${\displaystyle a+3b+12c=d}$
${\displaystyle 4a+2b+3c=d}$
I end up with 4 variables to find. What am I missing? Is there one more linear equation possible to add?

Let us have 3 vectors in ${\displaystyle {\mathbb{R}}^3}$:
${\displaystyle (a_1,b_1,c_1),(a_2,b_2,c_2)}$ and ${\displaystyle (a_3,b_3,c_3)}$
These vectors are linearly dependent if and only if the following scalar equation holds:
${\displaystyle a_1(b_2{c}_3-b_3{c}_2)-b_1(a_2{c}_3-a_3{c}_2)+c_1(a_2{b}_3-a_3{b}_2)=0}$
1. This equation is a necessary and sufficient condition of lineary dependence.
2. This criterion is a scalar equation (not a pair of equations; not a vector form).
3. This equation is a linear equation as to ${\displaystyle (a_i,b_i,c_i)}$ for all ${\displaystyle i\in \{1,2,3\}}$.
Now let us have 2 strings in ${\displaystyle {\mathbb{R}}^3}$:
${\displaystyle (a_1,b_1,c_1)}$ and ${\displaystyle (a_2,b_2,c_2)}$
Is there a similar way to determine whether this set of vectors in ${\displaystyle {\mathbb{R}}^3}$ is linearly dependent?
I.e. the way such as the following scalar equation:
${\displaystyle f(a_1,b_1,c_1,a_2,b_2,c_2)=0}$
so, that
1.This equation is a necessary and sufficient condition of linear dependence.
2.This criterion is a scalar equation (not a pair of equations; not a vector form).
3. This equation is a linear equation as to ${\displaystyle (a_i,b_i,c_i)}$ for all ${\displaystyle i\in \{1,2\}}$.

I mean, that whenever we talk about linear equations, we say that they always form a straight line on a graph. We make this sure by making a graph of some possible solutions to the equation and plotting the points on the graph. But can we make sure that all the solutions of the linear equation will always form a straight line because we cannot plot each and every solution of the equation on the graph because there are infinitely many solutions to a linear equation in 2 variables?

Could there be infinite solutions to a modular linear equation of the form
Ax = b mod n
when solving for x?

I am currently working through a chapter in my textbook, and it is on solving a form of linear differential equation.
I don't understand how ${\displaystyle \frac{d}{dt}{e}^{k\left(t\right)}\}x=0}$, since it says that ${\displaystyle \frac{d}{dt}{e}^{k\left(t\right)}=p\left(t\right)e^{k\left(t\right)}}$. Surely this would mean that ${\displaystyle \frac{d}{dt}{e}^{k\left(t\right)}x=x\cdot p\left(t\right)e^{k\left(t\right)}}$.

The slope of a linear equation of the form ${\displaystyle Ax+By+C=0}$ (general form) is given by $-\frac{A}{B}$. Could someone intuitively explain why we use this formula?
I've read that this is because the coefficient of x (coefficient of y) represent the "speed" at which x (y) increases and thus we can get the slope by dividing the "speed of x" by the "speed of y" - but I'm afraid it's not very intuitive to me.
Why do we use this formula and how is this equivalent to the more general slope formula,
${\displaystyle slope=-\frac{\text{change in y}}{\text{change in x}}}$

Let ${\displaystyle T:{\mathbb{R}}^2\to {\mathbb{R}}^3}$ be the linear transformation defined by ${\displaystyle T(x,y)=(y-x,-3x,-3y)}$
Write a linear equation defining the subspace ${\displaystyle Im\left(T\right)}$
${\displaystyle \cdots =0Z}$
(Write your answer in the form ${\displaystyle ax+by+cz}$. For example "${\displaystyle 2x+3y-4z}$")
So, I somehow learned how to find the image or kernels, but have no idea how to write a linear equation of it. Can someone help me with it? At least please give me an answer to this question I can figure out how to do it.

Recently came across to solve system of linear differential equation with the matrix method, I have learnt that the eigenvectors forms part of the complementary function.
A matrix method such as :
${\displaystyle \frac{dy}{dt}=5x+7y}$
${\displaystyle \frac{dx}{dt}=9x+4y}$
which then we replace a vector x =(x,y) and a matrix M with entries (57,94)
Is there any reason for this (such as the one in ordinary Differential equation is due to the fact that the exponential is the eigenfunction if the differential which also exist as a matrix method, but why the eigenvectors in the front of it?)

Help!
Find the equation of the line with slope = 8 and passing through (-8,-3). Write the equation in point-slope form AND slope-intercept forms. Include the full equation in your answers.
point-slope form:
slope-intercept form:

Is (9,5) a solution of the system of linear equations? ${\displaystyle -9x+y=-7}$ and ${\displaystyle 6x+y=4}$
a.Yes
b.No

Help me with solving the following system of linear equations by using Gauss-Seidel method:
${\displaystyle x_1-8x_2+3x_3=-9}$
${\displaystyle x_1-5x_2-6x_3=2}$
${\displaystyle x_1-3x_2+7x_3=-5}$
Where ${\displaystyle x_1^{\left(0\right)}=0.2,x{\left\{\left(0\right)\right\}}_2=0.6,x{\left\{\left(0\right)\right\}}_3=0.4}$, use one loop?

Help with solving the system of linear equations for all variables^
${\displaystyle 2x+6y-z=4}$
${\displaystyle 4x+5z=7}$
${\displaystyle 3y-2z=0}$

If we were given two points on a linear equation ${\displaystyle (x_1,y_1),(x_2,y_2)}$, it is quite easy to find the slope and use substitution to find the slope intercept form ${\displaystyle y=mx+b}$, to graph it.
Is it possible to solve for b strictly in terms of ${\displaystyle x_1,y_1,x_2,y_2}$?

The title might not be too descriptive; I'm not sure how to classify this question.
I have a set of linear equations of the form
${\displaystyle c_i{A}_i+p_i{B}_i=E}$
where ${\displaystyle A_i,B_i}$ and E are 3 by 1 column vectors and ${\displaystyle c_i}$ and ${\displaystyle p_i}$ are scalars. As the subscripts indicate, I have n such linear equations, where all the scalars and vectors except E change with each one. Another limitation is that all I know for each set of linear equations is ${\displaystyle A_i}$ and ${\displaystyle B_i}$, and I want to find E.
Is there a method for solving these equations?
The equation above is a simplification of the equation
${\displaystyle R_p^T{T}_D=c_i{A}_i-p_i{R}_D{B}_i}$
where ${\displaystyle R_p^T}$ is the inverse of a rotation matrix, ${\displaystyle T_D}$ is a translation vector and ${\displaystyle R_D}$ is another rotation matrix. I am interested in finding the rotation matrix ${\displaystyle R_p}$. Again, this equation describes the relationship between two homogenious points (${\displaystyle A_i}$ and ${\displaystyle B_i}$) in 3D space, where the z component is 1 (i.e., the c and p scalars project the points to somewhere in the 3D space other than (x, y, 1)). I have n point-pairs A and B.