### Video Transcript

Consider the function π¦ is equal to eight over the absolute value of two π₯ plus five minus three. Find a negative value of π₯ for which π¦ is equal to two.

We will begin by substituting π¦ is equal to two into our function. This gives us two is equal to eight divided by the absolute value of two π₯ plus five minus three. We can then multiply both sides of this equation by the denominator of the right-hand side. Next, we distribute our parentheses. We have two multiplied by the absolute value of two π₯ plus five minus six is equal to eight. Adding six to both sides of the equation gives us 14 on the right-hand side. Finally, we can divide both sides of this equation by two such that the absolute value or modulus of two π₯ plus five is equal to seven.

We can now solve this problem algebraically or graphically. We know that the absolute value or modulus of a real number is the nonnegative value of the number without regard to its sign. We sometimes consider this as the distance from zero. This means we will have two possible solutions to our equation. Either two π₯ plus five is equal to seven or two π₯ plus five is equal to negative seven. In both equations, we begin by subtracting five from both sides. This gives us two π₯ is equal to two and two π₯ is equal to negative 12. Dividing both sides of these equations by two gives us values of π₯ equal to one and negative six. Weβre asked to find a negative answer. Therefore, the correct answer is negative six.

As mentioned previously, we could also solve the equation the modulus of two π₯ plus five is equal to seven graphically. The expression inside the modulus function is written in the form ππ₯ plus π. This means it is a linear expression with slope π and π¦-intercept π. The linear equation π¦ is equal to two π₯ plus five can be drawn on the coordinate axes as shown. It intersects the π¦-axis at positive five and has a slope or gradient of two.

The modulus or absolute value of a function must always be positive. This means that to draw the equation π¦ is equal to the absolute value of two π₯ plus five, we reflect the portion of the graph below the π₯-axis so that the whole graph is a V shape above the π₯-axis. Weβre interested in when this graph is equal to seven. We can find our solutions by drawing a horizontal line at π¦ is equal to seven. This gives us two solutions, one of which is positive and one negative. The negative one will once again correspond to π₯ is equal to negative six.