Equations With Two Variables - PSAT Math
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What is the equation in standard form for the line $y=-3x+7$?
What is the equation in standard form for the line $y=-3x+7$?
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$3x+y=7$. Move $x$ term to left side.
$3x+y=7$. Move $x$ term to left side.
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Write the equation of the line with slope $3$ and $y$-intercept $-2$.
Write the equation of the line with slope $3$ and $y$-intercept $-2$.
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$y=3x-2$. Direct substitution into $y=mx+b$.
$y=3x-2$. Direct substitution into $y=mx+b$.
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Write the equation in slope-intercept form: $2x+y=7$.
Write the equation in slope-intercept form: $2x+y=7$.
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$y=-2x+7$. Isolate $y$ by subtracting $2x$ from both sides.
$y=-2x+7$. Isolate $y$ by subtracting $2x$ from both sides.
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Find the slope of the line through $(1,4)$ and $(3,0)$.
Find the slope of the line through $(1,4)$ and $(3,0)$.
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$m=-2$. Use $m=\frac{0-4}{3-1}=\frac{-4}{2}=-2$.
$m=-2$. Use $m=\frac{0-4}{3-1}=\frac{-4}{2}=-2$.
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Determine whether $(2,-1)$ is a solution to $3x+2y=4$.
Determine whether $(2,-1)$ is a solution to $3x+2y=4$.
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Yes, because $3(2)+2(-1)=4$. Substitute and check if equation holds.
Yes, because $3(2)+2(-1)=4$. Substitute and check if equation holds.
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Find the $y$-intercept of the line $3x+6y=12$.
Find the $y$-intercept of the line $3x+6y=12$.
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$\left(0,2\right)$. Set $x=0$: $6y=12$, so $y=2$.
$\left(0,2\right)$. Set $x=0$: $6y=12$, so $y=2$.
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Identify whether $(2,3)$ is a solution to $x+y=5$.
Identify whether $(2,3)$ is a solution to $x+y=5$.
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Yes. Check: $2+3=5$ ✓
Yes. Check: $2+3=5$ ✓
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Find the $x$-intercept of the line $3x+6y=12$.
Find the $x$-intercept of the line $3x+6y=12$.
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$\left(4,0\right)$. Set $y=0$: $3x=12$, so $x=4$.
$\left(4,0\right)$. Set $y=0$: $3x=12$, so $x=4$.
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Find the missing value $k$ so that $(3,k)$ satisfies $2x+y=11$.
Find the missing value $k$ so that $(3,k)$ satisfies $2x+y=11$.
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$5$. Substitute: $2(3)+k=11$, so $6+k=11$, thus $k=5$.
$5$. Substitute: $2(3)+k=11$, so $6+k=11$, thus $k=5$.
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Identify the equation of the vertical line through $x=-3$.
Identify the equation of the vertical line through $x=-3$.
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$x=-3$. Vertical lines have constant $x$-value for all $y$.
$x=-3$. Vertical lines have constant $x$-value for all $y$.
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What is the $x$-intercept of $Ax+By=C$ (assume $A\ne 0$)?
What is the $x$-intercept of $Ax+By=C$ (assume $A\ne 0$)?
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$\left(\frac{C}{A},0\right)$. Set $y=0$ and solve for $x$: $Ax=C$, so $x=\frac{C}{A}$.
$\left(\frac{C}{A},0\right)$. Set $y=0$ and solve for $x$: $Ax=C$, so $x=\frac{C}{A}$.
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What is the $y$-intercept of $Ax+By=C$ (assume $B\ne 0$)?
What is the $y$-intercept of $Ax+By=C$ (assume $B\ne 0$)?
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$\left(0,\frac{C}{B}\right)$. Set $x=0$ and solve for $y$: $By=C$, so $y=\frac{C}{B}$.
$\left(0,\frac{C}{B}\right)$. Set $x=0$ and solve for $y$: $By=C$, so $y=\frac{C}{B}$.
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What is the slope between points $(x_1,y_1)$ and $(x_2,y_2)$?
What is the slope between points $(x_1,y_1)$ and $(x_2,y_2)$?
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$m=\frac{y_2-y_1}{x_2-x_1}$. Rise over run: change in $y$ divided by change in $x$.
$m=\frac{y_2-y_1}{x_2-x_1}$. Rise over run: change in $y$ divided by change in $x$.
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What is the equation of the horizontal line through $y=-3$?
What is the equation of the horizontal line through $y=-3$?
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$y=-3$. Horizontal lines have zero slope and constant $y$-value.
$y=-3$. Horizontal lines have zero slope and constant $y$-value.
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What does it mean for $(x,y)$ to be a solution to an equation in two variables?
What does it mean for $(x,y)$ to be a solution to an equation in two variables?
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$(x,y)$ makes the equation true when substituted for $x$ and $y$. The point satisfies the equation when its coordinates are plugged in.
$(x,y)$ makes the equation true when substituted for $x$ and $y$. The point satisfies the equation when its coordinates are plugged in.
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Which equation is parallel to $y=\frac{1}{3}x-4$ and passes through $(0,2)$?
Which equation is parallel to $y=\frac{1}{3}x-4$ and passes through $(0,2)$?
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$y=\frac{1}{3}x+2$. Parallel lines share slope $\frac{1}{3}$; through $(0,2)$ gives $b=2$.
$y=\frac{1}{3}x+2$. Parallel lines share slope $\frac{1}{3}$; through $(0,2)$ gives $b=2$.
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Find the equation of the line with slope $-2$ passing through $(3,1)$.
Find the equation of the line with slope $-2$ passing through $(3,1)$.
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$y=-2x+7$. Use point-slope: $y-1=-2(x-3)$, simplify to slope-intercept form.
$y=-2x+7$. Use point-slope: $y-1=-2(x-3)$, simplify to slope-intercept form.
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Find the equation in slope-intercept form through $(0,2)$ and $(4,6)$.
Find the equation in slope-intercept form through $(0,2)$ and $(4,6)$.
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$y=x+2$. Slope $m=\frac{6-2}{4-0}=1$, passes through $(0,2)$ giving $b=2$.
$y=x+2$. Slope $m=\frac{6-2}{4-0}=1$, passes through $(0,2)$ giving $b=2$.
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Identify whether $(2,3)$ is a solution to $x+y=5$.
Identify whether $(2,3)$ is a solution to $x+y=5$.
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Yes, because $2+3=5$. Substitute $(2,3)$: $2+3=5$ is true, confirming it's a solution.
Yes, because $2+3=5$. Substitute $(2,3)$: $2+3=5$ is true, confirming it's a solution.
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Find the $y$-intercept of the line $3x-2y=12$.
Find the $y$-intercept of the line $3x-2y=12$.
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$\left(0,-6\right)$. Substitute $x=0$: $3(0)-2y=12$, so $-2y=12$, thus $y=-6$.
$\left(0,-6\right)$. Substitute $x=0$: $3(0)-2y=12$, so $-2y=12$, thus $y=-6$.
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Find the $x$-intercept of the line $3x-2y=12$.
Find the $x$-intercept of the line $3x-2y=12$.
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$\left(4,0\right)$. Substitute $y=0$: $3x-2(0)=12$, so $3x=12$, thus $x=4$.
$\left(4,0\right)$. Substitute $y=0$: $3x-2(0)=12$, so $3x=12$, thus $x=4$.
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Find the slope of the line $2x+5y=10$.
Find the slope of the line $2x+5y=10$.
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$m=-\frac{2}{5}$. Rewrite as $y=-\frac{2}{5}x+2$ to identify slope $m=-\frac{2}{5}$.
$m=-\frac{2}{5}$. Rewrite as $y=-\frac{2}{5}x+2$ to identify slope $m=-\frac{2}{5}$.
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Identify the slope and $y$-intercept in $y=-3x+5$.
Identify the slope and $y$-intercept in $y=-3x+5$.
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$m=-3$ and $b=5$. Read directly from $y=mx+b$ form: coefficient of $x$ and constant term.
$m=-3$ and $b=5$. Read directly from $y=mx+b$ form: coefficient of $x$ and constant term.
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What condition on slopes shows two nonvertical lines are perpendicular?
What condition on slopes shows two nonvertical lines are perpendicular?
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$m_1m_2=-1$. Perpendicular slopes multiply to $-1$ (negative reciprocals).
$m_1m_2=-1$. Perpendicular slopes multiply to $-1$ (negative reciprocals).
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What condition on slopes shows two nonvertical lines are parallel?
What condition on slopes shows two nonvertical lines are parallel?
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$m_1=m_2$. Parallel lines have same steepness, thus equal slopes.
$m_1=m_2$. Parallel lines have same steepness, thus equal slopes.
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What is the slope of the line through $(x_1,y_1)$ and $(x_2,y_2)$?
What is the slope of the line through $(x_1,y_1)$ and $(x_2,y_2)$?
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$m=\frac{y_2-y_1}{x_2-x_1}$. Rise over run formula calculates rate of change between two points.
$m=\frac{y_2-y_1}{x_2-x_1}$. Rise over run formula calculates rate of change between two points.
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What is the $y$-intercept of the line $Ax+By=C$ (assume $B\ne 0$)?
What is the $y$-intercept of the line $Ax+By=C$ (assume $B\ne 0$)?
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$\left(0,\frac{C}{B}\right)$. Set $x=0$ and solve for $y$ to find where line crosses $y$-axis.
$\left(0,\frac{C}{B}\right)$. Set $x=0$ and solve for $y$ to find where line crosses $y$-axis.
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What is the $x$-intercept of the line $Ax+By=C$ (assume $A\ne 0$)?
What is the $x$-intercept of the line $Ax+By=C$ (assume $A\ne 0$)?
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$\left(\frac{C}{A},0\right)$. Set $y=0$ and solve for $x$ to find where line crosses $x$-axis.
$\left(\frac{C}{A},0\right)$. Set $y=0$ and solve for $x$ to find where line crosses $x$-axis.
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What is the slope of the line given by $Ax+By=C$ in terms of $A$ and $B$?
What is the slope of the line given by $Ax+By=C$ in terms of $A$ and $B$?
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$m=-\frac{A}{B}$. Derived by solving $Ax+By=C$ for $y$ to get slope-intercept form.
$m=-\frac{A}{B}$. Derived by solving $Ax+By=C$ for $y$ to get slope-intercept form.
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What is the point-slope form of a line through $(x_1,y_1)$ with slope $m$?
What is the point-slope form of a line through $(x_1,y_1)$ with slope $m$?
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$y-y_1=m(x-x_1)$. Uses known point $(x_1,y_1)$ and slope $m$ to define the line.
$y-y_1=m(x-x_1)$. Uses known point $(x_1,y_1)$ and slope $m$ to define the line.
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