Question

Solve the following pair of equations
a) 2x+ 3y=40 , 2x+ 6y=60
b) 5x+ 2y=20 , 2x+3y=19

Answer 1

Answer:

In the first set of equation, $x=10\ and\ y=\frac{20}{3}\ and\ for\ the\ 2nd\ set\ of\ equations,\ x=2\ and\ y=5$

Step-by-step explanation:

The question is related to linear equation in one variable.

There are three ways in which this can be solved.

Well, I'll use only one method.

The pair of equations is

2x+3y=40 --(1)

2x+6y=60 --(2)

Now, simply subtract eq. 1 from eq. 2

$2x-2x+6y-3y=60-40\\ 3y=20\\ y=\frac{20}{3}\\ \therefore y=\frac{20}{3}\\ \\ putting\ value\ of\ y\ in\ eq.\ 1 \\ 2x+3\times \frac{20}{3}=40 \\ \\ \implies 2x+20=40\\ 2x=20 \\ \implies x=10$

Coming to the second question

The set of equations is

5x+ 2y=20 --(1)

2x+3y=19 --(2)

Now, multiply eq. (1) by 3 and eq. (2) by 2

15x+6y=60 --(3)

4x+6y=38 --(4)

Now, subtract eq. (4) by eq. (3)

$15x-4x+6y-6y=60-38\\ 11x=22\\ \implies \therefore x=2$

Putting x=2 in eq. (3)

(15×2)+6y=60

30+6y=60

6y=30

y=5

Answer 2

Answer:

(a) $2x+3y=40,\:2x+6y=60\quad :\quad y=\dfrac{20}{3},\:x=10$

(b) $5x+2y=20,\:2x+3y=19\quad :\quad y=5,\:x=2$

Step-by-step explanation:

(a)                    $\mathrm{Graph\:1}$

$\begin{bmatrix}2x+3y=40\\ 2x+6y=60\end{bmatrix}$

$\mathrm{Isolate}\:x\:\mathrm{for}\:2x+3y=40$

$2x+3y=40$

$\mathrm{Subtract\:}3y\mathrm{\:from\:both\:sides}$

$2x+3y-3y=40-3y$

$\mathrm{Divide\:both\:sides\:by\:}2$

$\dfrac{2x}{2}=\dfrac{40}{2}-\dfrac{3y}{2}$

$\mathrm{Simplify}$

$x=\dfrac{40-3y}{2}$

$\mathrm{Subsititute\:}x=\dfrac{40-3y}{2}$

$\begin{bmatrix}2\cdot \dfrac{40-3y}{2}+6y=60\end{bmatrix}$

$\mathrm{Isolate}\:y\:\mathrm{for}\:2\dfrac{40-3y}{2}+6y=60$

$2\cdot \dfrac{40-3y}{2}+6y=60$

$2\cdot \dfrac{40-3y}{2}$

$\mathrm{Multiply\:fractions}:\quad \:a\cdot \dfrac{b}{c}=\dfrac{a\:\cdot \:b}{c}$

$=\dfrac{\left(40-3y\right)\cdot \:2}{2}$

$\mathrm{Cancel\:the\:common\:factor:}\:2$

$=40-3y$

$40-3y+6y=60$

$\mathrm{Add\:similar\:elements:}\:-3y+6y=3y$

$40+3y=60$

$\mathrm{Subtract\:}40\mathrm{\:from\:both\:sides}$

$40+3y-40=60-40$

$\mathrm{Simplify}$

$3y=20$

$\mathrm{Divide\:both\:sides\:by\:}3$

$\dfrac{3y}{3}=\dfrac{20}{3}$

$\mathrm{Simplify}$

$y=\dfrac{20}{3}$

$\mathrm{For\:}x=\dfrac{40-3y}{2}$

$\mathrm{Subsititute\:}y=\dfrac{20}{3}$

$x=\dfrac{40-3\cdot \dfrac{20}{3}}{2}$

$\dfrac{40-3\cdot \dfrac{20}{3}}{2}$

$=\dfrac{40-20}{2}$

$\mathrm{Subtract\:the\:numbers:}\:40-20=20$

$=\dfrac{20}{2}$

$\mathrm{Divide\:the\:numbers:}\:\dfrac{20}{2}=10$

$x=10$

$\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}$

$y=\dfrac{20}{3},\:x=10$

(b)                    $\mathrm{Graph\:2}$

$\begin{bmatrix}5x+2y=20\\ 2x+3y=19\end{bmatrix}$

$\mathrm{Isolate}\:x\:\mathrm{for}\:5x+2y=20$

$\mathrm{Subsititute\:}x=\dfrac{20-2y}{5}$

$\begin{bmatrix}2\cdot \dfrac{20-2y}{5}+3y=19\end{bmatrix}$

$\mathrm{Isolate}\:y\:\mathrm{for}\:2\dfrac{20-2y}{5}+3y=19$

$2\cdot \dfrac{20-2y}{5}+3y=19$

$\mathrm{Expand\:}2\cdot \dfrac{20-2y}{5}+3y$

$2\cdot \dfrac{20-2y}{5}+3y$

$2\cdot \dfrac{20-2y}{5}$

$\mathrm{Multiply\:fractions}:\quad \:a\cdot \dfrac{b}{c}=\dfrac{a\:\cdot \:b}{c}$

$=\dfrac{\left(20-2y\right)\cdot \:2}{5}$

$\mathrm{Expand}\:\left(20-2y\right)\cdot \:2$

$=2\left(20-2y\right)$

$\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b-c\right)=ab-ac$

$a=2,\:b=20,\:c=2y$

$=2\cdot \:20-2\cdot \:2y$

$\mathrm{Simplify}\:2\cdot \:20-2\cdot \:2y$

$\mathrm{Multiply\:the\:numbers:}\:2\cdot \:20=40$

$=40-2\cdot \:2y$

$\mathrm{Multiply\:the\:numbers:}\:2\cdot \:2=4$

$=40-4y$

$=\dfrac{40-4y}{5}$

$\mathrm{Convert\:element\:to\:fraction}:\quad \:3y=\dfrac{3y5}{5}$

$=\dfrac{40-4y}{5}+\dfrac{3y\cdot \:5}{5}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \dfrac{a}{c}\pm \dfrac{b}{c}=\dfrac{a\pm \:b}{c}$

$=\dfrac{40-4y+3y\cdot \:5}{5}$

$40-4y+3y\cdot \:5=40+11y$

$=\dfrac{40+11y}{5}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \dfrac{a\pm \:b}{c}=\dfrac{a}{c}\pm \dfrac{b}{c}$

$\dfrac{40+11y}{5}=\dfrac{40}{5}+\dfrac{11y}{5}$

$=\dfrac{40}{5}+\dfrac{11y}{5}$

$\mathrm{Divide\:the\:numbers:}\:\dfrac{40}{5}=8$

$=8+\dfrac{11y}{5}$

$8+\dfrac{11y}{5}=19$

$\mathrm{Multiply\:both\:sides\:by\:}5$

$8\cdot \:5+\dfrac{11y}{5}\cdot \:5=19\cdot \:5$

$\mathrm{Simplify}$

$40+11y=95$

$\mathrm{Subtract\:}40\mathrm{\:from\:both\:sides}$

$40+11y-40=95-40$

$\mathrm{Simplify}$

$11y=55$

$\mathrm{Divide\:both\:sides\:by\:}11$

$\dfrac{11y}{11}=\dfrac{55}{11}$

$\mathrm{Simplify}$

$y=5$

$\mathrm{For\:}x=\dfrac{20-2y}{5}$

$\mathrm{Subsititute\:}y=5$

$x=\dfrac{20-2\cdot \:5}{5}$

$\dfrac{20-2\cdot \:5}{5}$

$=\dfrac{10}{5}$

$\mathrm{Divide\:the\:numbers:}\:\dfrac{10}{5}=2$

$=2$

$x=2$

$\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}$

$y=5,\:x=2$