+2446 Let's go through the process again, shall we?
Let x represent one of the numbers. Then 15-x is the other number. Knowing this, the equation becomes \(x(15-x)=63\).
| \(x(15-x)=63\) | Distribute. |
| \(15x-x^2=63\) | Subtract 63 from both sides. |
| \(-x^2+15x-63=0\) | I like to divide by -1 out of habit. |
| \(x^2-15x+63=0\) | In this case, this quadratic cannot be factored. I'll use the quadratic formula here. |
| \(x=\frac{-(-15)\pm\sqrt{(-15)^2-4(1)(63)}}{2(1)}\) | Let's do some simplification here. |
| \(x=\frac{15\pm\sqrt{225-252}}{2}\) | |
| \(x=\frac{15\pm\sqrt{-27}}{2}\) | I see an issue here, do you? This is probably why you were unable to come up with a solution. The square root of -27 results in a nonreal answer. |
This means that no quadratic equation with real solutions can have property aforementioned, which is a sum of 15 and a product of 63.
+2446 This one requires some thinking. We know that a quadratic equation has 2 zeroes. One is located at \((x,0)\), and the other is located at \((y,0)\).
Now, what do we know about the relationship of these two numbers? Well, we know that the sum of both zeroes is -15.
| \(x+y=-15\) | Let's solve for y here. |
| \(y=-15-x\) | |
Since y=-15-x, we know that a zero is located at \((-15-x,0)\). We also know that, when multiplied, both zeroes yield -54. Knowing this information, we can generate a quadratic equation.
\(x(-15-x)=-54\)
This quadratic equation has everything that we are looking for. Now, solve for the zeroes of this quadratic.
| \(x(-15-x)=-54\) | Distribute the x into the binomial in the parentheses. | ||
| \(-15x-x^2=-54\) | Add 54 to both sides. | ||
| \(-x^2-15x+54=0\) | Divide by -1. This is not strictly necessary, but it does make the x^2-term positive, which could make methods of solving easier. | ||
| \(x^2+15x-54=0\) | Factor this by finding two numbers that multiply to -54 and add to 15. These happen to be -3 and 18. | ||
| \((x-3)(x+18)=0\) | By the zero product theorem, set both products equal to zero. | ||
| Add the constant in both equations and move it move to the right hand side. | ||
| |||
These zeroes definitely fit the criteria.
+2446 Evaluating the expression of \(\frac{\frac{3}{8}}{1\frac{7}{8}}-\frac{1}{6}\)
| \(\frac{\frac{3}{8}}{1\frac{7}{8}}-\frac{1}{6}\) | First, convert the fraction in the denominator into an improper fraction. |
| \(\frac{\frac{3}{8}}{1\frac{7}{8}}-\frac{1}{6}=\frac{\frac{3}{8}}{\frac{8*1+7}{8}}-\frac{1}{6}=\frac{\frac{3}{8}}{\frac{15}{8}}-\frac{1}{6}\) | Multiply by the reciprocal of the denominator to eliminate it. |
| \(\frac{\frac{8}{15}}{\frac{8}{15}}*\frac{\frac{3}{8}}{\frac{15}{8}}-\frac{1}{6}\) | Doing this isn't actually changing the value of the fraction because we are just multiplying by 1. |
| \(\frac{8}{15}*\frac{3}{8}-\frac{1}{6}\) | Before beginning the multiplication, we can drastically simplify the numbers in both fractions by identifying the GCF of opposite numerators and denominators. In this example, 8 and 8 have a GCF of 8. 3 and 15 have a GCF if 3. |
| \(\frac{1}{5}-\frac{1}{6}\) | Convert 1/5 and 1/6 into fractions with a common denominator. |
| \(\frac{6}{30}-\frac{5}{30}\) | Now subtract the numerators while maintaining the denominator. |
| \(\frac{1}{30}\) | |
+2446 The table of x- and y-values definitely represent a linear equation because the increase from one point to another is constant. This is a direct indicator of a linear function.
If you cannot figure it by trial and error, then you can do this:
\(y=mx+b\)
We know that all linear functions are in this form. Let's plug in a point into this equation. I'll use the first one in the table, (2,8)
| \(y=mx+b\) | Plug in 2 for x and 8 for y. |
| \(8=2m+b\) | |
Let's do this again for a separate point. In this case, I'll use (3,11). It does not matter which point you choose. The only criteria is that it must be different from the one you previously chose.
| \(y=mx+b\) | Plug in the coordinate (3,11) into the equation. |
| \(11=3m+b\) | |
Look at this! We have a system of equations that we can solve for both m and b. I'll use substitution for this demonstration because it is easier to showcase on an online format. I'll solve for b in the first equation, 8=2m+b
| \(8=2m+b\) | Subtracting 2m from both sides is the easiest method to isolate one variable. |
| \(8-2m=b\) | |
Now, plug 8-2m in for b in the second equation.
| \(11=3m+b\) | Plug in the aforementioned value for b. |
| \(11=3m+8-2m\) | Simplify the right hand side of the equation. |
| \(11=m+8\) | Subtract 8 on both sides. |
| \(m=3\) | |
Now, substitute m=3 into either original equation. I'll do 8=2m+b
| \(8=2m+b\) | Plug in 3 for m. |
| \(8=6+b\) | Subtract 6 from both sides. |
| \(b=2\) | |
Therefore, the equation that satisfies that is \(y=3x+2\). You can confirm this by testing out each coordinate. Of course, it works.
+2446 This may help, if multiplying by 9 mentally is difficult. A lot of these mental tricks I wish I knew earlier on.
| \(15*9\) | Let's make it simpler. |
| \(15(10-1)\) | Obviously, this is the same thing. Now, distribute the 15 to both terms. |
| \(15*10-15*1\) | Now, simplify. This is much easier, don't you think? |
| \(150-15\) | And just in case borrowing is another mental headache... |
| \(150-(20-5)\) | Yet again, I have not changed the expression at all. |
| \(150-20+5\) | Simplify. |
| \(130+5\) | |
| \(135\) | |
This may look like a lot of steps, but all we are doing is making the problem easier. With practice, this will simplify multiplication with large numbers.
