How to solve systems of equations with multiplication & addition.

$a+b=12$

$a\cdot b=36$

$a+b=12$

$a\cdot b=36$

Raul Walker
2022-07-15
Answered

How to solve systems of equations with multiplication & addition.

$a+b=12$

$a\cdot b=36$

$a+b=12$

$a\cdot b=36$

You can still ask an expert for help

Alexia Hart

Answered 2022-07-16
Author has **19** answers

There are two ways to solve a system of equations, the first one being substitution and other one being elimination. In this case, we will use substitution which requires us to put one variable in terms of another. Currently we have the following equations:

$\begin{array}{rl}& a+b=12\\ & a\cdot b=36\end{array}$

Since we have two variables and as I said before, we have to rearrange one in terms of another. We can either choose $a$ or $b$. We'll choose $a$. If rearrange for $a$ we get:

$a=-b+12$

Since,we have said that $a=b-12$, we can substitute $a$ into our second equation so we get:

$(-b+12)(b)=36$

Now we can expand and see that we have a quadratic equation:

$-{b}^{2}+12b=36$

Now we bring $36$ to the left side and make equation equal to $0$.

$-{b}^{2}-12b-36=0$

There is no need to get complex here but for future equations, you should check if the equation has $1$, $2$, or no real roots through the discriminant ${b}^{2}-4ac$. If the value is below zero then it has no real roots, if it is above zero than it has $2$ real roots and if it is equal to zero then it has $1$ root. In this case, the discriminant is equal to zero and therefore, has only one root.

You can use whatever method you like to solve equation but the answer that you will get is $6$.

$\begin{array}{rl}& a+b=12\\ & a\cdot b=36\end{array}$

Since we have two variables and as I said before, we have to rearrange one in terms of another. We can either choose $a$ or $b$. We'll choose $a$. If rearrange for $a$ we get:

$a=-b+12$

Since,we have said that $a=b-12$, we can substitute $a$ into our second equation so we get:

$(-b+12)(b)=36$

Now we can expand and see that we have a quadratic equation:

$-{b}^{2}+12b=36$

Now we bring $36$ to the left side and make equation equal to $0$.

$-{b}^{2}-12b-36=0$

There is no need to get complex here but for future equations, you should check if the equation has $1$, $2$, or no real roots through the discriminant ${b}^{2}-4ac$. If the value is below zero then it has no real roots, if it is above zero than it has $2$ real roots and if it is equal to zero then it has $1$ root. In this case, the discriminant is equal to zero and therefore, has only one root.

You can use whatever method you like to solve equation but the answer that you will get is $6$.

civilnogwu

Answered 2022-07-17
Author has **6** answers

If you are given numbers $r$ and $s$, and you have to find numbers $a$ and $b$ such that

$a+b=r,\phantom{\rule{2em}{0ex}}ab=s$

then you can think about the polynomial

$(x-a)(x-b)={x}^{2}-(a+b)x+ab={x}^{2}-rx+s$

So, the solutions of the quadratic equation

${x}^{2}-rx+s=0$

are the numbers $a$ and $b$ that you want. Can you solve a quadratic equation?

$a+b=r,\phantom{\rule{2em}{0ex}}ab=s$

then you can think about the polynomial

$(x-a)(x-b)={x}^{2}-(a+b)x+ab={x}^{2}-rx+s$

So, the solutions of the quadratic equation

${x}^{2}-rx+s=0$

are the numbers $a$ and $b$ that you want. Can you solve a quadratic equation?

asked 2022-06-23

Two equations & three unknowns (in$\mathbb{Z}$)

this system-equation has answer $(x,y,z)$ in Integers Set or not?

${a}_{1}x+{b}_{1}y+{c}_{1}z={d}_{1}$

this system-equation has answer $(x,y,z)$ in Integers Set or not?

${a}_{1}x+{b}_{1}y+{c}_{1}z={d}_{1}$

asked 2022-06-19

Find the solution set of triplets $(x,y,z)$ that fulfil this system using Gauss-Jordan:

$\{\begin{array}{l}-x+2z=0\\ 3x-6z=0\\ 2x-4z=0\end{array}$

First of all, I don't see any $y$ variable there. I suppose it doesn't matter and I proceed normally:

$\left[\begin{array}{ccc}-1& 2& 0\\ 3& -6& 0\\ 2& -4& 0\end{array}\right]$

$-{f}_{1}$

$\left[\begin{array}{ccc}1& -2& 0\\ 3& -6& 0\\ 2& -4& 0\end{array}\right]$

$-3{f}_{1}+{f}_{2}$

$\left[\begin{array}{ccc}1& -2& 0\\ 0& 0& 0\\ 2& -4& 0\end{array}\right]$

$-2{f}_{1}+{f}_{3}$

$\left[\begin{array}{ccc}1& -2& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

So, this is the staggered reduced form.

This is an homogeneous system (because of the null column), thus one solution is $(0,0,0)$.

Other than that, I have to check out the range of the system. The range is $1$, which is less than the number of columns... what is the number of columns?

$\{\begin{array}{l}-x+2z=0\\ 3x-6z=0\\ 2x-4z=0\end{array}$

First of all, I don't see any $y$ variable there. I suppose it doesn't matter and I proceed normally:

$\left[\begin{array}{ccc}-1& 2& 0\\ 3& -6& 0\\ 2& -4& 0\end{array}\right]$

$-{f}_{1}$

$\left[\begin{array}{ccc}1& -2& 0\\ 3& -6& 0\\ 2& -4& 0\end{array}\right]$

$-3{f}_{1}+{f}_{2}$

$\left[\begin{array}{ccc}1& -2& 0\\ 0& 0& 0\\ 2& -4& 0\end{array}\right]$

$-2{f}_{1}+{f}_{3}$

$\left[\begin{array}{ccc}1& -2& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

So, this is the staggered reduced form.

This is an homogeneous system (because of the null column), thus one solution is $(0,0,0)$.

Other than that, I have to check out the range of the system. The range is $1$, which is less than the number of columns... what is the number of columns?

asked 2022-09-11

Is there any simple analytic method for solving $\sqrt{x}+y=7$ and $x+\sqrt{y}=11$ simultaneously.

asked 2022-07-07

Determining Whether a System of Linear Equations Is Inconsistent

1- Is getting a false equation during the normal process of solving a system of linear equations (e.g. substitution/elimination) is the only way we ensure this system has no solutions?

2- If no, what other ways available?

1- Is getting a false equation during the normal process of solving a system of linear equations (e.g. substitution/elimination) is the only way we ensure this system has no solutions?

2- If no, what other ways available?

asked 2022-11-18

Solving system of equations: ${x}^{3}-3{y}^{2}x=-1$ and $3y{x}^{2}-{y}^{3}=1$

asked 2022-06-22

How many solutions are there to this equation?

$\begin{array}{rl}{x}^{2}-{y}^{2}& =z\\ {y}^{2}-{z}^{2}& =x\\ {z}^{2}-{x}^{2}& =y\end{array}$

$\begin{array}{rl}{x}^{2}-{y}^{2}& =z\\ {y}^{2}-{z}^{2}& =x\\ {z}^{2}-{x}^{2}& =y\end{array}$

asked 2022-10-12

Solve the following linear system:

$$\begin{array}{}-a{\pi}_{0}+(1-b){\pi}_{1}=0\\ a{\pi}_{0}-{\pi}_{1}+{\pi}_{2}=0\\ b{\pi}_{1}-{\pi}_{2}=0\\ {\pi}_{0}+{\pi}_{1}+{\pi}_{2}=1\end{array}$$

Here, ${\pi}_{0},\phantom{\rule{thinmathspace}{0ex}}{\pi}_{1},\phantom{\rule{thinmathspace}{0ex}}{\pi}_{2}$ are the variables.

$$\begin{array}{}-a{\pi}_{0}+(1-b){\pi}_{1}=0\\ a{\pi}_{0}-{\pi}_{1}+{\pi}_{2}=0\\ b{\pi}_{1}-{\pi}_{2}=0\\ {\pi}_{0}+{\pi}_{1}+{\pi}_{2}=1\end{array}$$

Here, ${\pi}_{0},\phantom{\rule{thinmathspace}{0ex}}{\pi}_{1},\phantom{\rule{thinmathspace}{0ex}}{\pi}_{2}$ are the variables.