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3 years ago

Using substitution, what is the first step to solve: x=6y+3 and x+2y=5.

Mathematics

2 answers:

Sedbober [7]3 years ago

7 0

Answer:

x= 9/2 y=1,4

Step-by-step explanation:

Hope this helped you! :)

Fiesta28 [93]3 years ago

4 0

Answer:

Step-by-step explanation:

Remark

You could do it a very convoluted way or you could do it simply. The simple way, I think, is the one intended. The x on the second equation is sitting by it itself. So is it in the first equation. Put the x from the first equation into the second.

Solution

x=6y+3 ===> x+2y=5.

6y + 3 + 2y = 5 First Step

8y + 3 = 5 Subtract 3 from both sides

8y = 5 - 3 Combine

8y = 2 Divide by 8

y = 2/8

y = 1/4

============================

x = 6y + 3

x = 6*(1/4) + 3

x = 6/4 + 3

x = 1.5 + 3

x = 4.5

Solution (4.5 , 0.25)

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Classify each number according to its value.

Tems11 [23]

Answer:See below and attached

Step-by-step explanation:

Given numbers

4.2×10^-6, 2.1×10^-3, 3.1×10^-2, 3.2×10^-5, 3.5×10^-4, 5.8×10^-3, 5.2×10^-4

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3.1×10^-2, 5.8×10^-3

Between 3.1 × 10^-3 and 4.3 × 10^-5

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Less than 4.3 × 10^-5

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Step-by-step explanation

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2 years ago

Each of the 14 students in the art club needs 4 ounces of paint for a project. The art store sells paint only in 8-ounce bottles

Agata [3.3K]

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3 years ago

Read 2 more answers

Show work please<br> \sqrt(x+12)-\sqrt(2x+1)=1

Nesterboy [21]

Answer:

$x=4$

Step-by-step explanation:

Given $\displaystyle\\\sqrt{x+12}-\sqrt{2x+1}=1$ , start by squaring both sides to work towards isolating $x$ :

$\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2$

Recall $(a-b)^2=a^2-2ab+b^2$ and $\sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}$ :

$\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2\\\implies x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1$

Isolate the radical:

$\displaystyle\\x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1\\\implies -2\sqrt{(x+12)(2x+1)}=-3x-12\\\implies \sqrt{(x+12)(2x+1)}=\frac{-3x-12}{-2}$

Square both sides:

$\displaystyle\\(x+12)(2x+1)=\left(\frac{-3x-12}{-2}\right)^2$

Expand using FOIL and $(a+b)^2=a^2+2ab+b^2$ :

$\displaystyle\\2x^2+25x+12=\frac{9}{4}x^2+18x+36$

Move everything to one side to get a quadratic:

$\displaystyle-\frac{1}{4}x^2+7x-24=0$

Solving using the quadratic formula:

A quadratic in $ax^2+bx+c$ has real solutions $\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ . In $\displaystyle-\frac{1}{4}x^2+7x-24$ , assign values:

$\displaystyle \\a=-\frac{1}{4}\\b=7\\c=-24$

Solving yields:

$\displaystyle\\x=\frac{-7\pm \sqrt{7^2-4\left(-\frac{1}{4}\right)\left(-24\right)}}{2\left(-\frac{1}{4}\right)}\\\\x=\frac{-7\pm \sqrt{25}}{-\frac{1}{2}}\\\\\begin{cases}x=\frac{-7+5}{-0.5}=\frac{-2}{-0.5}=\boxed{4}\\x=\frac{-7-5}{-0.5}=\frac{-12}{-0.5}=24 \:(\text{Extraneous})\end{cases}$