qanda-logo
search-icon
Symbol
apple-logoApp Storegoogle-play-logoGoogle Play

Calculator search results

Solve the system of equations
Answer
circle-check-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
expand-arrow-icon
Graph
$0.6 x - 1.3 y = - 1.5$
$\dfrac { 1 } { 4 } x - \dfrac { 2 } { 9 } y = \dfrac { 1 } { 3 }$
$x$Intercept
$\left ( - \dfrac { 5 } { 2 } , 0 \right )$
$y$Intercept
$\left ( 0 , \dfrac { 15 } { 13 } \right )$
$x$Intercept
$\left ( \dfrac { 4 } { 3 } , 0 \right )$
$y$Intercept
$\left ( 0 , - \dfrac { 3 } { 2 } \right )$
$x = 4 , y = 3$
Solve the system of equations
$\begin{cases} \color{#FF6800}{ 0.6 } \color{#FF6800}{ x } - 1.3 y = - 1.5 \\ \dfrac { 1 } { 4 } x - \dfrac { 2 } { 9 } y = \dfrac { 1 } { 3 } \end{cases}$
$ $ Calculate the multiplication expression $ $
$\begin{cases} \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } \color{#FF6800}{ x } } { \color{#FF6800}{ 5 } } } - 1.3 y = - 1.5 \\ \dfrac { 1 } { 4 } x - \dfrac { 2 } { 9 } y = \dfrac { 1 } { 3 } \end{cases}$
$\begin{cases} \dfrac { 3 x } { 5 } \color{#FF6800}{ - } \color{#FF6800}{ 1.3 } \color{#FF6800}{ y } = - 1.5 \\ \dfrac { 1 } { 4 } x - \dfrac { 2 } { 9 } y = \dfrac { 1 } { 3 } \end{cases}$
$ $ Calculate the multiplication expression $ $
$\begin{cases} \dfrac { 3 x } { 5 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 13 } \color{#FF6800}{ y } } { \color{#FF6800}{ 10 } } } = - 1.5 \\ \dfrac { 1 } { 4 } x - \dfrac { 2 } { 9 } y = \dfrac { 1 } { 3 } \end{cases}$
$\begin{cases} \dfrac { 3 x } { 5 } - \dfrac { 13 y } { 10 } = \color{#FF6800}{ - } \color{#FF6800}{ 1.5 } \\ \dfrac { 1 } { 4 } x - \dfrac { 2 } { 9 } y = \dfrac { 1 } { 3 } \end{cases}$
$ $ Convert decimals to fractions $ $
$\begin{cases} \dfrac { 3 x } { 5 } - \dfrac { 13 y } { 10 } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 2 } } } \\ \dfrac { 1 } { 4 } x - \dfrac { 2 } { 9 } y = \dfrac { 1 } { 3 } \end{cases}$
$\begin{cases} \dfrac { 3 x } { 5 } - \dfrac { 13 y } { 10 } = - \dfrac { 3 } { 2 } \\ \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 4 } } } \color{#FF6800}{ x } - \dfrac { 2 } { 9 } y = \dfrac { 1 } { 3 } \end{cases}$
$ $ Calculate the multiplication expression $ $
$\begin{cases} \dfrac { 3 x } { 5 } - \dfrac { 13 y } { 10 } = - \dfrac { 3 } { 2 } \\ \color{#FF6800}{ \dfrac { \color{#FF6800}{ x } } { \color{#FF6800}{ 4 } } } - \dfrac { 2 } { 9 } y = \dfrac { 1 } { 3 } \end{cases}$
$\begin{cases} \dfrac { 3 x } { 5 } - \dfrac { 13 y } { 10 } = - \dfrac { 3 } { 2 } \\ \dfrac { x } { 4 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 9 } } } \color{#FF6800}{ y } = \dfrac { 1 } { 3 } \end{cases}$
$ $ Calculate the multiplication expression $ $
$\begin{cases} \dfrac { 3 x } { 5 } - \dfrac { 13 y } { 10 } = - \dfrac { 3 } { 2 } \\ \dfrac { x } { 4 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ y } } { \color{#FF6800}{ 9 } } } = \dfrac { 1 } { 3 } \end{cases}$
$\begin{cases} \dfrac { 3 x } { 5 } - \dfrac { 13 y } { 10 } = - \dfrac { 3 } { 2 } \\ \dfrac { x } { 4 } - \dfrac { 2 y } { 9 } = \dfrac { 1 } { 3 } \end{cases}$
$ $ Solve a solution to $ x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 13 } } { \color{#FF6800}{ 6 } } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } } \\ \dfrac { x } { 4 } - \dfrac { 2 y } { 9 } = \dfrac { 1 } { 3 } \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 13 } } { \color{#FF6800}{ 6 } } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ \dfrac { \color{#FF6800}{ x } } { \color{#FF6800}{ 4 } } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ y } } { \color{#FF6800}{ 9 } } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 3 } } } \end{cases}$
$ $ Substitute the given $ x $ value into the equation $ \dfrac { x } { 4 } - \dfrac { 2 y } { 9 } = \dfrac { 1 } { 3 }$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ \dfrac { \color{#FF6800}{ 13 } } { \color{#FF6800}{ 6 } } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } } } { \color{#FF6800}{ 4 } } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ y } } { \color{#FF6800}{ 9 } } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 3 } } }$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ \dfrac { \color{#FF6800}{ 13 } } { \color{#FF6800}{ 6 } } } \color{#FF6800}{ y } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } } } { \color{#FF6800}{ 4 } } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } \color{#FF6800}{ y } } { \color{#FF6800}{ 9 } } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 3 } } }$
$ $ Solve a solution to $ y$
$\color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$ $ Substitute the given $ y $ value into the equation $ x = \dfrac { 13 } { 6 } y - \dfrac { 5 } { 2 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 13 } } { \color{#FF6800}{ 6 } } } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } }$
$x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 13 } } { \color{#FF6800}{ 6 } } } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } - \dfrac { 5 } { 2 }$
$ $ Calculate the product of rational numbers $ $
$x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 13 } } { \color{#FF6800}{ 2 } } } - \dfrac { 5 } { 2 }$
$x = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 13 } } { \color{#FF6800}{ 2 } } } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 5 } } { \color{#FF6800}{ 2 } } }$
$ $ Find the difference between the two fractions $ \dfrac { 13 } { 2 } $ and $ - \dfrac { 5 } { 2 }$
$x = \color{#FF6800}{ 4 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 }$
$ $ The possible solutions are as follows $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
$ $ Check if it is the solution to the system of equations $ $
$\begin{cases} \color{#FF6800}{ 0.6 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 1.3 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ - } \color{#FF6800}{ 1.5 } \\ \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 4 } } } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 9 } } } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 3 } } } \end{cases}$
$\begin{cases} \color{#FF6800}{ 0.6 } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ 1.3 } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ - } \color{#FF6800}{ 1.5 } \\ \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 4 } } } \color{#FF6800}{ \times } \color{#FF6800}{ 4 } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 2 } } { \color{#FF6800}{ 9 } } } \color{#FF6800}{ \times } \color{#FF6800}{ 3 } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 3 } } } \end{cases}$
$ $ Simplify the equality $ $
$\begin{cases} \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 2 } } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 3 } } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 3 } } } \end{cases}$
$\begin{cases} \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 2 } } } = \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 3 } } { \color{#FF6800}{ 2 } } } \\ \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 3 } } } = \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 3 } } } \end{cases}$
$ $ Since it is true in both equations, it is the solution of the system of equations $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ 4 } , \color{#FF6800}{ y } = \color{#FF6800}{ 3 }$
Solution search results
search-thumbnail-Which of the following rational numbers are
Which of the following rational numbers are equivalent? $0Ptionsy$ A \frac{5}{6}, \frac{30}{36} B $s\sqrt{rac\left(} -2\right)\left(3\right)\sqrt{1rac} \sqrt{4\right)16\right)4} $ C $s\sqrt{11aC\left(} -4\right)1-7b,\sqrt{1rac\left(16\sqrt{35\right)9} } $ D \frac{1}{2},\frac{3}{8}
7th-9th grade
Other
Search count: 5,909
Check solutionright-arrow-icon
search-thumbnail-(int|imits -0a1(1-\timesn_{2}{)_{3}}^{n}
$\left(int|imits$ $-0a1\left(1-\times n_{2}{\right)_{3}}^{n}$ $x^{A}3dx=7\right)$ $\left($ $frac\left(1\right)\left(40\right)\right)$ $\left($ $\left(troc\left(1\right)\left(35\right)\right)$ $\left(troc\left(1\right)\left(30\right)\right)$ $\left(tr0c\left(1\right)\left(25\right)\right)$
Calculus
Check solutionright-arrow-icon
search-thumbnail-In the following problem a,b, b,and represent REAL NUMBERS.
In the following problem $a,b,$ b,and c represent REAL NUMBERS. The derivative of $g\left(x\right)=log _{a}\left(bx\right)+cx^{2}is$ $○$ $g^{1}\left(x\right)=\right)dfrac\left(b\right)log _{-}a\left(bx\right)\right)\left(N1n$ $a\right)+2cx1\right)$ $○$ $\left(g\left(x\right)=\right)dtracb$ $loga\left(bx\right)+2c\times \right)Nln$ a}\) $○$ $g^{'}\left(x\right)=\left(dfraC\left(a\right)log _{-}a\left(bx\right)\right)\left(ln$ $\right)+2c\times 1\right)\right)$ $○$ $g^{1}\left(x\right)=\left(dfrac\left(b$ $log _{-}a\left(bx\right)\right)\left(ln$ $a1\right)$ $○$ $\left(\left(g^{1}\left(x\right)=b$ $log _{-}a\left(bx\right)+2cx1\right)$ $○$ $g\left(x\right)=\left(dfrac\left(b$ $log _{-}a\left(bx\right)\right)\left(1ln$ $a|+2c1\right)$
Calculus
Check solutionright-arrow-icon
search-thumbnail-$The\ reciprocal\\ 0+11) \left(\frac{2}
$8$ $\left(1$ Point) $1\right)$ The\ reciprocal\\ $0+11\right)$ \left(\frac{2} $c\left(2\right)$ {5}\right)^0\ $\right)$ \ $1111s\right)$ $S$ $S1S$ $s3S$ $S4S$ $s2S$
7th-9th grade
Other
Search count: 4,895
Check solutionright-arrow-icon
search-thumbnail-(10) 0.2x+0.3y=1.3
$\left(10\right)$ $0.2x+0.3y=1.3$ $0.4x+05y=23$ $3x$ $51$
10th-13th grade
Other
Check solutionright-arrow-icon
search-thumbnail-( limlimits_{Degin{array}{I) x ightarrow 1) \y ightarrow 1 end{array)}}
$\left($ limlimits_{Degin{array}{I) $x$ ightarrow $1\right)$ \y ightarrow $1$ end{array)}} $ac\left(\times \times 2-y^{n}$ $2\right)\times \left(x-y\right)=7\right)$
Algebra
Search count: 969
Check solutionright-arrow-icon
Have you found the solution you wanted?
Try again
Try more features at Qanda!
check-iconSearch by problem image
check-iconAsk 1:1 question to TOP class teachers
check-iconAI recommend problems and video lecture
apple-logoApp Storegoogle-play-logoGoogle Play
© 2021 Mathpresso Inc.
|
CEO Jongheun Lee, Yongjae Lee
|
17th Floor, WeWork Seolleung Station III, 428, Seolleung-ro, Gangnam-gu, Seoul
|
EMAIL support.en@mathpresso.com