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kramer
2 years ago
8

Solve the following using Substitution method 2x – 5y = -13 3x + 4y = 15

Mathematics
1 answer:
Digiron [165]2 years ago
8 0

\huge \boxed{\mathfrak{Question} \downarrow}

Solve the following using Substitution method

2x – 5y = -13

3x + 4y = 15

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

\left. \begin{array}  { l  }  { 2 x - 5 y = - 13 } \\ { 3 x + 4 y = 15 } \end{array} \right.

  • To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

2x-5y=-13, \: 3x+4y=15

  • Choose one of the equations and solve it for x by isolating x on the left-hand side of the equal sign. I'm choosing the 1st equation for now.

2x-5y=-13

  • Add 5y to both sides of the equation.

2x=5y-13

  • Divide both sides by 2.

x=\frac{1}{2}\left(5y-13\right)  \\

  • Multiply \frac{1}{2}\\ times 5y - 13.

x=\frac{5}{2}y-\frac{13}{2}  \\

  • Substitute \frac{5y-13}{2}\\ for x in the other equation, 3x + 4y = 15.

3\left(\frac{5}{2}y-\frac{13}{2}\right)+4y=15  \\

  • Multiply 3 times \frac{5y-13}{2}\\.

\frac{15}{2}y-\frac{39}{2}+4y=15  \\

  • Add \frac{15y}{2} \\ to 4y.

\frac{23}{2}y-\frac{39}{2}=15  \\

  • Add \frac{39}{2}\\ to both sides of the equation.

\frac{23}{2}y=\frac{69}{2}  \\

  • Divide both sides of the equation by 23/2, which is the same as multiplying both sides by the reciprocal of the fraction.

\large \underline{ \underline{ \sf \: y=3 }}

  • Substitute 3 for y in x=\frac{5}{2}y-\frac{13}{2}\\. Because the resulting equation contains only one variable, you can solve for x directly.

x=\frac{5}{2}\times 3-\frac{13}{2}  \\

  • Multiply 5/2 times 3.

x=\frac{15-13}{2}  \\

  • Add -\frac{13}{2}\\ to \frac{15}{2}\\ by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.

\large\underline{ \underline{ \sf \: x=1 }}

  • The system is now solved. The value of x & y will be 1 & 3 respectively.

\huge\boxed{  \boxed{\bf \: x=1, \: y=3 }}

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The volume of this cube is 27 cubic meters. What is the value of m?
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Hey ! there

Answer:

  • n is equal to <u>3 </u><u>meters</u>

Step-by-step explanation:

In this question we are provided with a cube having <u>volume </u><u>2</u><u>7</u><u> </u><u>cubic </u><u>meters</u><u> </u>. And we are asked to find the <u>value </u><u>of </u><u>n </u>that is basically its <u>edge </u><u>.</u>

For finding the value of n we need to know the volume of cube. So ,

\:  \qquad  \:  \qquad \: \underline{ \boxed{ \frak{Volume_{(Cube)} = a {}^{3} }}}

<u>Where</u><u> </u><u>,</u>

  • a refers to <u>edge </u><u>of </u><u>cube</u>

<u>SOLUTION</u><u> </u><u>:</u><u> </u><u>-</u>

Substituting given volume that is 27 m³ and value of a as n in formula :

\quad \longrightarrow \qquad \: n {}^{3}  = 27

Applying cube root on both sides :

\quad \longrightarrow \qquad \:  \sqrt[3]{n {}^{3} }  =  \sqrt[3]{27}

We get ,

\quad \longrightarrow \qquad \:n=  \sqrt[3]{27}

We know that 3 × 3 × 3 is equal to 27 that means cube root of 27 is 3 . So ,

\quad \longrightarrow \qquad \:    \blue{\underline{\boxed{\frak{ n = 3 \: m}}}} \quad \bigstar

  • <u>Henceforth</u><u> </u><u>,</u><u> </u><u>value</u><u> </u><u>of </u><u>n </u><u>is </u><u>❝</u><u> </u><u>3 </u><u>meters </u><u>❞</u>

<u>Verifying</u><u> </u><u>:</u><u> </u><u>-</u>

Now we are checking our answer whether it is wrong or right by substituting value of n and equating it with given volume that is 27 cubic meters . So ,

  • a³ = 27 ( where a is equal to n )

Substituting values :

  • ( 3 )³ = 27

  • 3 × 3 × 3 = 27

  • 9 × 3 = 27

  • 27 = 27

  • L.H.S = R.H.S

  • Hence, Verified .

<u>Therefore,</u><u> </u><u>our</u><u> answer</u><u> is</u><u> correct</u><u> </u><u>.</u>

<h2><u>#</u><u>K</u><u>e</u><u>e</u><u>p</u><u> </u><u>Learning</u></h2>
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