Evaluate the algebraic expression when x = − 3 x=-3 x = − 3 and y = 2 y=2 y = 2 . x ( 20 − 15 y ) − ∣ 2 x + y ∣ x\left(20-15y\right)-\left|2x+y\right| x ( 20 − 15 y ) − ∣ 2 x + y ∣

Welcome back, everyone. So in this practice problem, we're gonna evaluate this algebraic expression. I've got x and then 20 minus 15 y in parenthesis minus and I've got 2x+y, but it's all in an absolute value sign. But the idea is the same. I'm just gonna replace x's with negative 3, y's with 2. So let's go ahead and get started. So these x's will become negative 3, and these y's will become 2. So if I do this, what happens is I get negative 3 on the outside, and then I get 20 minus 15. And then what happens is because I have a y that's attached to a 15 that indicates multiplication, so I have to do a parenthesis. Otherwise, it would look like a 152, and that would be weird. So then we have minus, absolute value, and I've got 2. And, again, I'm doing parenthesis. I'm doing multiplication, so I wanna add 2 times negative 3. And then over here now, I've got plus y. Now because I'm just adding a y that isn't being multiplied, I don't have to put the parentheses. So I can just rule replace it with a 2. But if you wanted to, you could include a little parenthesis. That's totally fine. Alright? So now we're basically done with the replacing. All we have to do here is now we have to just stick to our order of operations. Remember, the first thing you do is parentheses, but notice how we also have absolute value signs. And absolute value signs basically behave like parentheses, so it's always a good idea to do your your, parentheses and your absolute values together, because whatever happens inside of this absolute value sign, that's just going to become positive. Alright? So we're gonna focus on these two expressions first. So what we're gonna do here is we're gonna do negative 3, right, nothing changes there, and then we have to do whatever's inside the parentheses. And in this parentheses, we have subtraction and multiplication. So it's weird because we have a parentheses that also includes some of the other sort of later operations, but remember, multiplication always comes first. So what we're gonna do here is we're gonna do 20 minuteus, and then we have to do 15 times 2. You always have to do the multiplication first, so that's 20 minuteus 30. And over here when what happens to the absolute value is, again, we've got multiplication and then addition over here. We have to do the multiplication first. So this just becomes negative 6 plus, and this becomes 2. Alright? So now let's simplify a little bit further. So this becomes negative 3, and then this over here, this 20 minuteus 30 becomes negative 10. Right? That's what that becomes. Then we have minus and then absolute value, negative 6 plus 2 this becomes negative 4. Alright? So now let's go ahead and do this here. Just be really careful because you have a negative sign that's distributing into another negative sign, so negative three times negative ten now just becomes 30. And the reason we can do this is because we've basically done the parentheses and the absolute values. So now we can go ahead and, you know, there's no exponents, so now we can just go into multiplication and division before addition and subtraction. So the multiplication here of negative three times negative ten becomes positive 30, and then over here what happens is we have 30 minuteus, and then this absolute value really just becomes positive 4. Just be careful because you might be tempted to distribute a minus sign because you see a minus and then another minus, but, technically, you won't be able to do that always because, remember, just this thing here becomes just positive 4. So this just becomes 4 over here. That's what this becomes. Alright? And then now what happens is now you've done the multiplication, and now your last thing is to do addition and subtraction, now that you've sort of done the, absolute value. So this becomes 30 minus 4, which is 26, and that should be your final answer. Alright? So, that's it for this one. Thanks for watching.

0. Review of Algebra

Algebraic Expressions

College Algebra

0. Review of Algebra

Algebraic Expressions

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Multiple Choice

Evaluate the algebraic expression when $x=-3$ and $y=2$ .
$x$\left$(20-15y$\right$)-$\left$|2x+y$\right$|$

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Verified step by step guidance

Substitute the given values of x and y into the expression. Replace x with -3 and y with 2 in the expression x(20 - 15y) - |2x + y|.

Calculate the expression inside the parentheses: 20 - 15y. Substitute y = 2 to get 20 - 15(2).

Simplify the expression 20 - 15(2) to find its value.

Calculate the absolute value expression |2x + y|. Substitute x = -3 and y = 2 to get |2(-3) + 2|.

Simplify the expression 2(-3) + 2 and then find the absolute value of the result.

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Evaluate the algebraic expression when x=−3x=-3x=−3 and y=2y=2y=2.x(20−15y)−∣2x+y∣x\(\left\)(20-15y\(\right\))-\(\left\)|2x+y\(\right\)|x(20−15y)−∣2x+y∣

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Algebraic Expressions practice set

Evaluate the algebraic expression when $x=-3$ and $y=2$ .
$x\(\left\)(20-15y\(\right\))-\(\left\)|2x+y\(\right\)|$