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3 years ago

2x+y=3, 5x-2y=12 how to solve this

2 answers:

7 0

Start by getting one of the equations in terms of one variable, in this case lets put the first equation in terms of y

2x+y=3 ---> y=3-2x

now we plug that into the second equation

5x-2y=12 ---> 5x-2(3-2x)=12

we can then multiply and simplify

5x-2(3-2x)=12 ---> 5x-6+4x=12 ---> 9x-6=12 ---> 9x=18 ---> x=2

we can now plug x=2 into an equation to solve for y

2(2)+y=3 ---> 4+y=3 ---> y=-1

Brums [2.3K]3 years ago

7 0

2x+y=3 =4x+2y=6
5x-2y=12. =5x-2y=12
--------------------------------
9x=18: x=2
4+y=3:y=-1

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Drag the tiles to the boxes to form correct pairs. Match the pairs of equivalent expressions.

Rashid [163]

Answer:

The following pairs/results are matched:

$5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$
$3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Step-by-step explanation:

Lets solve all the expressions to match the results.

$5\left(2t+1\right)+\left(-7t+28\right)$

Solving the expression

$5\left(2t+1\right)+\left(-7t+28\right)$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$5\left(2t+1\right)-7t+28$

$10t+5-7t+28$

$3t+33$

Therefore, $5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$

$3\left(3t-4\right)-\left(2t+10\right)$

Solving the expression

$3\left(3t-4\right)-\left(2t+10\right)$

$9t-12-\left(2t+10\right)$

$9t-12-2t-10$

$7t-22$

Therefore, $3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$

Solving the expression

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$4t-\frac{8}{5}-\left(3-\frac{4}{3}t\right)$

$4t-\frac{8}{5}-\left(-\frac{4t}{3}+3\right)$

$4t-\frac{8}{5}-3+\frac{4t}{3}$

$-3-\frac{8}{5}:\quad -\frac{23}{5}$ and $\frac{4t}{3}+4t:\quad \frac{16t}{3}$

So,

$4t-\frac{8}{5}-3+\frac{4t}{3}$ will become $\frac{16t}{3}-\frac{23}{5}$

Therefore, $\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$

Solving the expression

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$-\frac{9}{2}t+3+\frac{7}{4}t+33$

$\mathrm{Group\:like\:terms}$

$\frac{9}{2}t+\frac{7}{4}t+3+33$

$\mathrm{Add\:similar\:elements:}\:-\frac{9}{2}t+\frac{7}{4}t=-\frac{11}{4}t$

$-\frac{11}{4}t+3+33$

$-\frac{11}{4}t+36$

Therefore, $\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Thus, the following pairs/results are matched:

$5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$
$3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Keywords: algebraic expression

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3 years ago

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Mrs. Ness buys 22 oz. of bananas for $2.86. What was the cost per ounce of the bananas?

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Answer:

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Step-by-step explanation:

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3 years ago

Pls help i really really need help on this. We just learned this but i can't remember it.

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Answer:

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Step-by-step explanation:

Area of shaded region

= area of circle -area of square

$\boxed{area \: of \: circle = \pi {r}^{2} }$

Given that radius= 10cm, r= 10cm.

$\boxed{area \: of \: square = side \: \times side}$

Let the length of a side of the square be x cm.

Applying Pythagoras Theorem,

10² +10²= x²

100 +100= x²

x²= 200

$x = \sqrt{200}$

Area of square

= x²

= 200cm²

Area of shaded region

= 3.14(10²) -200

= 3.14(100) -200

= 314 -200

= 114 cm²

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3 years ago

How do you solve x to the second power +7x=O

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