Question

$z=15x^2+10xy-6x-4y$ For which of the ordered pairs, $(x,y),$ below is $z\ne 0?$ a. (-3,2) b. (-2,3) c. $(\frac{2}{5},0)$ d. $(\frac{2}{5},10)$

Short answer

The ordered pair $(-3,2)$ makes $zeq0$.

Step-by-step solution

- Understand the problem

Given a function of two variables, $z=15x^2+10xy-6x-4y$, determine for which of the provided ordered pairs $(x,y)$, the value of $z$ is not equal to zero.

- Substitute pair (-3, 2)

Substitute $x=-3$ and $y=2$ into the equation. $$z=15(-3{)}^2+10(-3)(2)-6(-3)-4(2)$$ $$z=15(9)+10(-6)+18-8$$ $$z=135-60+18-8$$ $$z=85$$Since $85eq0$, the pair $(-3,2)$ makes $zeq0$.

- Substitute pair (-2, 3)

Substitute $x=-2$ and $y=3$ into the equation. $$z=15(-2{)}^2+10(-2)(3)-6(-2)-4(3)$$ $$z=15(4)+10(-6)+12-12$$ $$z=60-60+12-12$$ $$z=0$$Since $0=0$, the pair $(-2,3)$ does not make $zeq0$.

- Substitute pair $(\frac{2}{5},0)$

Substitute $x=\frac{2}{5}$ and $y=0$ into the equation. $$z=15{\left(\frac{2}{5}\right)}^2+10\left(\frac{2}{5}\right)(0)-6\left(\frac{2}{5}\right)-4(0)$$ $$z=15\left(\frac{4}{25}\right)+0-\frac{12}{5}+0$$ $$z=\frac{60}{25}-\frac{12}{5}$$ $$z=\frac{12}{5}-\frac{12}{5}$$ $$z=0$$ Since $0=0$, the pair $(\frac{2}{5},0)$ does not make $zeq0$.

- Substitute pair $(\frac{2}{5},10)$

Substitute $x=\frac{2}{5}$ and $y=10$ into the equation. $$z=15{\left(\frac{2}{5}\right)}^2+10\left(\frac{2}{5}\right)(10)-6\left(\frac{2}{5}\right)-4(10)$$ $$z=15\left(\frac{4}{25}\right)+10(4)-\frac{12}{5}-40$$ $$z=\frac{60}{25}+40-\frac{12}{5}-40$$ $$z=2.4+40-2.4-40$$ $$z=0$$ Since $0=0$, the pair $(\frac{2}{5},10)$ does not make $zeq0$.

Key concepts

Variable Substitution

Variable substitution is a crucial technique in solving mathematical problems, especially in algebra. This concept involves replacing variables in an equation with given numerical values.
For example, in the given problem, we have the function of two variables: $z=15x^2+10xy-6x-4y$. To check which ordered pair results in $zeq0$, we substitute the values of $x$ and $y$ from each pair into the equation:

• For the pair $(x,y)=(-3,2)$, we substitute $x=-3$ and $y=2$ into the equation.
• For the pair $(x,y)=(-2,3)$, substitute $x=-2$ and $y=3$.
• For the pair $(x,y)=(\frac{2}{5},0)$, substitute $x=\frac{2}{5}$ and $y=0$.
• For the pair $(x,y)=(\frac{2}{5},10)$, substitute $x=\frac{2}{5}$ and $y=10$.

With each substitution, you replace the variables in the function, then simplify to check if $z$ results in a non-zero value.

Quadratic Equations

Quadratic equations are algebraic expressions where the highest power of the variable is squared. In this problem, the term $15x^2$ indicates that we are dealing with a quadratic equation in one of the variables.
When substituting values, you can notice how the squared term affects the overall value of $z$:
Formula example with substitution:
For $(x,y)=(-3,2)$, the term $15x^2$ becomes $15(-3{)}^2=135$. This significant value has a large impact on $z$.

Due to the squared term, small changes in $x$ can cause large changes in $z$, highlighting the importance of careful substitution and simplification when dealing with quadratic components of an equation.

Algebraic Manipulations

Algebraic manipulations refer to the processes of rearranging and simplifying expressions. In our problem, after substituting the values into the equation, we perform several steps to simplify the expressions:

• Expand any squared terms, like $(-3{)}^2=9$.
• Multiply constants and variables, such as $15(9)$ and $10(-6)$.
• Combine like terms, which means adding or subtracting terms with the same variables.

For example, in Step 2, the calculation process is:
$$15(-3{)}^2=13510(-3)(2)=-60-6(-3)=184(2)=8Combiningthesegives:135-60+18-8=85$$

This manipulation shows how each component of the equation contributes to the final value of $z$.

Problem-Solving Techniques

Effective problem-solving techniques involve understanding the problem, systematic substitution, careful simplification, and evaluation of results. Here's how you can approach this type of problem:

• First, interpret the problem to understand what is required – in this case, finding when $zeq0$.
• Next, substitute the given pairs $(x,y)$ into the function one at a time.
• Simplify the resulting equations through careful algebraic manipulations and combine like terms.
• Finally, evaluate whether the simplified result is zero or not, confirming if the pair makes $zeq0$.

These steps not only help solve specific problems but also build a strong foundation in handling various algebraic challenges, fostering a deeper understanding of algebraic manipulations and quadratic equations.