// Numbas version: finer_feedback_settings {"name": "Year 11 Algebra quiz - Testing", "metadata": {"description": "

This quiz contains questions on algebraic fractions, logarithmic equations, exponential equations, quadratic equations and simultaneous equations.

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Evaluate \\(f(\\var{a3})\\)

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\\(f(\\var{a3})\\) = [[0]]

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Evaluating a function

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\\(f(x)=\\var{a1}x^{\\var{a2}}+\\var{b1}x-\\var{c1}\\)

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\\(x=\\var{a3}\\)

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\\(f(\\var{a3})=\\var{a1}*(\\var{a3})^{\\var{a2}}+\\var{b1}*(\\var{a3})-\\var{c1}\\)

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\\(f(\\var{a3})=\\simplify{{a1}*{a3}^{{a2}}}+\\simplify{{b1}*{a3}}-\\var{c1}\\)

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\\(f(\\var{a3})=\\simplify{{a1}*{a3}^{{a2}}+{b1}*{a3}-{c1}}\\)

", "statement": "

Given the function:

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\\(f(x)=\\var{a1}x^{\\var{a2}}+\\var{b1}x-\\var{c1}\\)

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Manipulation of algebraic fractions

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Express your answer as a fraction:

\n

               \\(V =\\) [[0]]

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When one fraction equals another fraction we can clear both fractions by cross-multiplying:

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\\((\\var{a}V+1)*(\\var{d}R+7)=(\\var{b}R+3)*(\\var{c}V+5)\\)

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\\(\\simplify{{a}*{d}}VR+\\simplify{7*{a}}V+\\var{d}R+7=\\simplify{{b}*{c}}VR+\\simplify{5*{b}}R+\\simplify{3*{c}}V+15\\)

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Gathering all the terms involving \\(V\\) to the left hand side and moving all other terms to the right hand side gives

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\\(\\simplify{{a}*{d}-{b}*{c}}VR+\\simplify{7*{a}-3*{c}}V=\\simplify{5*{b}-{d}}R+8\\)

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Factoring \\(V\\) out on the left hand side 

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\\(V(\\simplify{{a}*{d}-{b}*{c}}R+\\simplify{7*{a}-3*{c}})=\\simplify{5*{b}-{d}}R+8\\)

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Thus

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\\(V=\\frac{\\simplify{5*{b}-{d}}R+8}{\\simplify{{a}*{d}-{b}*{c}}R+\\simplify{7*{a}-3*{c}}}\\)

", "statement": "

Rearrange the following expression to make V the subject:

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     \\(\\frac{\\var{a}V+1}{\\var{b}R+3}=\\frac{\\var{c}V+5}{\\var{d}R+7}\\)

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Manipulation of an exponential function

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\\(x =\\) [[0]]

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\\(y=\\var{k}(1-\\var{c}e^{\\var{m}x+\\var{d}})\\)

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Working from the outside in, we divide across by \\(\\var{k}\\)   

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\\(\\frac{y}{\\var{k}}=1-\\var{c}e^{\\var{m}x+\\var{d}}\\)

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We can bring the \\(x\\) variable to the left hand side and move the \\(y\\) variable to the right hand side

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\\(\\var{c}e^{\\var{m}x+\\var{d}}=1-\\frac{y}{\\var{k}}\\)

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Again working from the outside in we divide across by \\(\\var{c}\\)

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\\(e^{\\var{m}x+\\var{d}}=\\frac{1-\\frac{y}{\\var{k}}}{\\var{c}}\\)

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Taking the natural log of both sides eliminates the \\(e\\) from the left hand side. 

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\\(\\var{m}x+\\var{d}=ln\\left(\\frac{1-\\frac{y}{\\var{k}}}{\\var{c}}\\right)\\)

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Subtract \\(\\var{d}\\) from both sides

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\\(\\var{m}x=ln\\left(\\frac{1-\\frac{y}{\\var{k}}}{\\var{c}}\\right)-\\var{d}\\)

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and finally divide by \\(\\var{m}\\) to get

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\\(x=\\frac{ln\\left(\\frac{1-\\frac{y}{\\var{k}}}{\\var{c}}\\right)-\\var{d}}{\\var{m}}\\)

", "statement": "

Rearrange the following expression to make \\(x\\) the subject:

\n

               \\(y=\\var{k}(1-\\var{c}e^{\\var{m}x+\\var{d}})\\)

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Calculate the value of \\(x\\) that satisfies the equation when  \\(y=\\var{d}\\).

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Input your answer correct to three decimal places.

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\\(x = \\) [[0]]

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Given the following logarithmic equation:

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\\(y=\\var{a}log(\\var{b}x+\\var{c}))\\)

\n

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Solve a logarithmic equation

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\\(\\var{a}log(\\var{b}x+\\var{c})=\\var{d}\\)

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Divide across by \\(\\var{a}\\)

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\\(log(\\var{b}x+\\var{c})=\\var{d}/\\var{a}=\\simplify{{d}/{a}}\\)

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\\(\\var{b}x+\\var{c}=10^{\\simplify{{d}/{a}}}\\)

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\\(\\var{b}x+\\var{c}=\\simplify{10^{{d}/{a}}}\\)

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\\(\\var{b}x=\\simplify{10^{{d}/{a}}}-\\var{c}\\)

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\\(\\var{b}x=\\simplify{10^{{d}/{a}}-{c}}\\)

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\\(x=\\simplify{(10^{{d}/{a}}-{c})/{b}}\\)

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\\(\\var{k}=\\var{a}e^{\\var{m}x+{\\var{c}}}\\)

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\\(\\frac{\\var{k}}{\\var{a}}=e^{\\var{m}x+\\var{c}}\\)

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\\(ln\\left(\\frac{\\var{k}}{\\var{a}}\\right)=\\var{m}x+\\var{c}\\)

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\\(ln\\left(\\frac{\\var{k}}{\\var{a}}\\right)-\\var{c}=\\var{m}x\\)

\n

\\(\\frac{ln\\left(\\frac{\\var{k}}{\\var{a}}\\right)-\\var{c}}{\\var{m}}=x\\)

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Solve an exponential equation

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Given the equation \\(f(x)=\\var{a}e^{\\var{m}x+\\var{c}}\\)

\n

Determine the value for \\(x\\) that satisfies the relation \\(f(x)=\\var{k}\\)

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Input your answer correct to three decimal places.

\n

\\(x = \\) [[0]]

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There are two values that satisfy the quadratic equation:

\n

\\(\\var{a1}x^2+\\simplify{{{a1}*{b1}*{c1}}}=\\simplify{{a1}*{b1}+{a1}{c1}}x\\)

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Type in the greater of the two values that satisfies the equation.

\n

Input your answer correct to three decimal places.  \\(x = \\) [[0]]

\n

Type in the lesser of the two values that satisfies the equation. 

\n

Input your answer correct to three decimal places.  \\(x = \\) [[1]]

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Solving quadratic equations using a formula,

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The formula for solving a quadratic equation of the form  \\(ax^2+bx+c=0\\)  is given by

\n

\\(x=\\frac{-b\\pm \\sqrt{b^2-4ac}}{2a}\\)

\n

In this example  \\(a=\\var{a1},\\,\\,\\,b=\\simplify{+-{a1}*({b1}+{c1})}\\)  and  \\(c=\\simplify{{a1}*{b1}*{c1}}\\)

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\\(x=\\frac{\\var{b}\\pm \\sqrt{(-\\var{b})^2-4*\\var{a1}*\\var{c}}}{2*\\var{a1}}\\)

\n

\\(x=\\frac{\\var{b}\\pm \\sqrt{\\simplify{{b}^2-4*{a1}*{c}}}}{\\simplify{2*{a1}}}\\)

\n

\\(x=\\simplify{{b}+({b}^2-4*{a1}*{c})^0.5}/\\simplify{2*{a1}}=\\simplify{({b}+({b}^2-4*{a1}*{c})^0.5)/(2*{a1})}\\)        or        \\(x=\\simplify{{b}-({b}^2-4*{a1}*{c})^0.5}/\\simplify{2*{a1}}=\\simplify{({b}-({b}^2-4*{a1}*{c})^0.5)/(2*{a1})}\\)

\n

\n

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Type in the greater of the two values that satisfies the equation. Input your answer correct to three decimal places.

\n

\\(x\\) = [[0]]

\n

Type in the lesser of the two values that satisfies the equation. Input your answer correct to three decimal places.

\n

\\(x\\) = [[1]]

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There are two values that satisfy the quadratic function below when  \\(y=\\var{c1}\\):

\n

\\(y=\\var{a1}x^2+\\var{b1}x\\)

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Solving quadratic equations using a formula,

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The formula for solving a quadratic equation of the form  \\(ax^2+bx+c=0\\)  is given by

\n

\\(x=\\frac{-b\\pm \\sqrt{b^2-4ac}}{2a}\\)

\n

In this example  \\(a=\\var{a1},\\,\\,\\,b=\\var{b1}\\)  and  \\(c=\\var{c1}\\)

\n

\\(x=\\frac{-\\var{b1}\\pm \\sqrt{\\var{b1}^2-4\\times\\var{a1}\\times\\var{c1}}}{2\\times\\var{a1}}\\)

\n

\\(x=\\frac{-\\var{b1}\\pm \\sqrt{\\simplify{{b1}^2-4*{a1}*{c1}}}}{\\simplify{2*{a1}}}\\)

\n

\\(x=\\simplify{(-{b1}+ ({b1}^2-4*{a1}*{c1})^0.5)/(2*{a1})}\\)   or   \\(x=\\simplify{(-{b1}- ({b1}^2-4*{a1}*{c1})^0.5)/(2*{a1})}\\)

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The following equation can be converted into a quadratic equation:

\n

\\(\\var{a1}x+\\frac{\\simplify{{a1}*{b1}*{c1}}}{x}=\\simplify{{a1}*({b1}+{c1})}\\)

", "showQuestionGroupNames": false, "tags": [], "variable_groups": [], "variablesTest": {"maxRuns": "1", "condition": ""}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Solving quadratic equations using a formula

"}, "advice": "

\\(\\var{a1}x+\\frac{\\simplify{{a1}*{b1}*{c1}}}{x}=\\simplify{{a1}*({b1}+{c1})}\\)

\n

We clear the fraction in the equation by multiplying across by \\(x\\)

\n

\\(\\var{a1}x^2+\\simplify{{a1}*{b1}*{c1}}=\\simplify{{a1}*({b1}+{c1})}x\\)

\n

Bringing all the terms to the left hand side and putting them in order of their powers of \\(x\\) gives

\n

\\(\\var{a1}x^2-\\simplify{{a1}*({b1}+{c1})}x+\\simplify{{a1}*{b1}*{c1}}=0\\)

\n

The formula for solving a quadratic equation of the form  \\(ax^2+bx+c=0\\)  is given by

\n

\\(x=\\frac{-b\\pm \\sqrt{b^2-4ac}}{2a}\\)

\n

In this example  \\(a=\\var{a1},\\,\\,\\,b=\\simplify{+-{a1}*({b1}+{c1})}\\)  and  \\(c=\\simplify{{a1}*{b1}*{c1}}\\)

\n

\\(x=\\frac{\\var{b}\\pm \\sqrt{(-\\var{b})^2-4*\\var{a1}*\\var{c}}}{2*\\var{a1}}\\)

\n

\\(x=\\frac{\\var{b}\\pm \\sqrt{\\simplify{{b}^2-4*{a1}*{c}}}}{\\simplify{2*{a1}}}\\)

\n

\\(x=\\simplify{{b}+({b}^2-4*{a1}*{c})^0.5}/\\simplify{2*{a1}}=\\simplify{({b}+({b}^2-4*{a1}*{c})^0.5)/(2*{a1})}\\)        or        \\(x=\\simplify{{b}-({b}^2-4*{a1}*{c})^0.5}/\\simplify{2*{a1}}=\\simplify{({b}-({b}^2-4*{a1}*{c})^0.5)/(2*{a1})}\\)

\n

\n

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Type in the greater of the two values that satisfies the equation. 

\n

\\(x = \\) [[0]]

\n

Type in the lesser of the two values that satisfies the equation. 

\n

\\(x = \\) [[1]]

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Input the value for \\(x\\) as an exact fraction.

\n

\\(x = \\) [[0]]

\n

Input the value for \\(y\\) as an exact fraction.

\n

\\(y = \\) [[1]]

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Solve the following system of simultaneous equations:

\n

\\(\\var{a}x+\\var{b}y=\\var{r1}\\)

\n

and

\n

\\(\\var{c}x+\\var{d}y=\\var{r2}\\)

", "rulesets": {}, "variable_groups": [], "ungrouped_variables": ["a", "b", "c", "d", "r1", "r2"], "functions": {}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Solving two simultaneous linear equations

"}, "variablesTest": {"condition": "{a}*{d}>{b}*{c}\n", "maxRuns": 100}, "advice": "

equation (i)    \\(\\var{a}x+\\var{b}y=\\var{r1}\\)

\n

equation (ii)    \\(\\var{c}x+\\var{d}y=\\var{r2}\\)

\n

If we decide to eliminate the \\(x\\) variables we need to have the same number of \\(x\\) in both equations

\n

\\(\\var{c}\\)*equation (i)      \\(\\simplify{{c}*{a}}x+\\simplify{{c}*{b}}y=\\simplify{{c}*{r1}}\\)

\n

\\(\\var{a}\\)*equation (ii)     \\(\\simplify{{c}*{a}}x+\\simplify{{d}*{a}}y=\\simplify{{a}*{r2}}\\)

\n

Subtracting gives:

\n

                      \\(\\simplify{{c}*{b}-{d}*{a}}y=\\simplify{{c}*{r1}-{a}*{r2}}\\)

\n

                       \\(y=\\simplify{({c}*{r1}-{a}*{r2})/({c}*{b}-{d}*{a})}\\)

\n

Substituting this solution for \\(y\\) into equation (i) gives

\n

       \\(\\var{a}x+\\var{b}*(\\simplify{({c}*{r1}-{a}*{r2})/({c}*{b}-{d}*{a})})=\\var{r1}\\)

\n

        \\(\\var{a}x=\\var{r1}-\\var{b}*(\\simplify{({c}*{r1}-{a}*{r2})/({c}*{b}-{d}*{a})})\\)

\n

        \\(\\var{a}x=\\simplify{{r1}-{b}*({c}*{r1}-{a}*{r2})/({c}*{b}-{d}*{a})}\\)

\n

\n

        \\(x=\\simplify{({r1}-{b}*({c}*{r1}-{a}*{r2})/({c}*{b}-{d}*{a}))/{a}}\\)

\n

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Solve a system of three simultaneous linear equations

"}, "parts": [{"type": "gapfill", "prompt": "

Input the value of \\(x\\) that satisfies the three equations.

\n

\\(x = \\) [[0]]

\n

Input the value of \\(y\\) that satisfies the three equations.

\n

\\(y = \\) [[1]]

\n

Input the value of \\(z\\) that satisfies the three equations.

\n

\\(z = \\) [[2]]

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(i)    \\(\\var{a1}x+2y+4z=\\var{r1}\\)

\n

(ii)   \\(2x+\\var{b1}y+3z=\\var{r2}\\)

\n

(iii)  \\(5x+6y+\\var{c1}z=\\var{r3}\\)

\n

First reduce the three equations in three unknowns to a two equations in two unknowns problem by eliminating one of the variables.

\n

We can eliminate \\(x\\) using equations (i) and (ii)

\n

2*(i)     \\(\\simplify{2*{a1}}x+4y+8z=\\simplify{2*{r1}}\\)

\n

\\(\\var{a1}\\)*(ii)    \\(\\simplify{2*{a1}}x+\\simplify{{a1}*{b1}}y+\\simplify{3*{a1}}z=\\simplify{{a1}*{r2}}\\)

\n

Subtracting gives us a new equation

\n

(iv)    \\(\\simplify{(4-{a1}{b1})y+(8-3*{a1})z}=\\simplify{2*{r1}-{a1}*{r2}}\\)

\n

We can also eliminate \\(x\\) using equations (ii) and (iii)

\n

5*(ii)    \\(10x +\\simplify{5*{b1}}y+15z=\\simplify{5*{r2}}\\)

\n

2*(iii)   \\(10x+12y+\\simplify{2*{c1}}z=\\simplify{2*{r3}}\\)

\n

Subtracting gives us another new equation

\n

(v)     \\(\\simplify{(5*{b1}-12)y+(15-2*{c1})z}=\\simplify{5*{r2}-2*{r3}}\\)

\n

We could then eliminate the \\(y\\) from these two new equations

\n

\\(\\simplify{5*{b1}-12}\\)*(iv)    \\(\\simplify{(5*{b1}-12)*(4-{a1}{b1})y+(5*{b1}-12)*(8-3*{a1})z}=\\simplify{(5*{b1}-12)*(2*{r1}-{a1}*{r2})}\\)

\n

\\(\\simplify{4-{a1}{b1}}\\)*(v)    \\(\\simplify{(4-{a1}{b1})*(5*{b1}-12)y+(4-{a1}{b1})*(15-2*{c1})z}=\\simplify{(4-{a1}{b1})*(5*{r2}-2*{r3})}\\)

\n

Subtracting gives us

\n

\\(\\simplify{(5*{b1}-12)*(8-3*{a1})-(4-{a1}{b1})*(15-2*{c1})}z=\\simplify{(5*{b1}-12)*(2*{r1}-{a1}*{r2})-(4-{a1}{b1})*(5*{r2}-2*{r3})}\\)

\n

Thus

\n

\\(z=\\frac{\\simplify{(5*{b1}-12)*(2*{r1}-{a1}*{r2})-(4-{a1}{b1})*(5*{r2}-2*{r3})}}{\\simplify{(5*{b1}-12)*(8-3*{a1})-(4-{a1}{b1})*(15-2*{c1})}}=\\simplify{decimal{((5*{b1}-12)*(2*{r1}-{a1}*{r2})-(4-{a1}*{b1})*(5*{r2}-2*{r3}))/(  (5*{b1}-12)*(8-3*{a1})-(4-{a1}*{b1})*(15-2*{c1}))}}\\)

\n

We can now back substitute this value for \\(z\\) into equation (iv) to find the correct value for \\(y\\) and then back substitute both these values into equation (i) to calculate \\(x\\). 

\n

", "statement": "

Solve the following system of three simultaneous linear equations:

\n

\\(\\var{a1}x+2y+4z=\\var{r1}\\)

\n

and

\n

\\(2x+\\var{b1}y+3z=\\var{r2}\\)

\n

and

\n

\\(5x+6y+\\var{c1}z=\\var{r3}\\)

", "type": "question"}, {"name": "Solving a Linear and a Non-linear system of equations", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}], "metadata": {"description": "

Solving a Linear and a Non-linear system of equations

", "licence": "Creative Commons Attribution 4.0 International"}, "rulesets": {}, "statement": "

Given two equations:

\n

\\(\\var{a1}x+\\var{b1}y=\\var{r1}\\)

\n

and

\n

\\(\\var{c1}x^2+\\var{d1}y^2=\\var{r2}\\)

\n

There are two solutions for \\(x\\) that satisfy both of these equations and for each \\(x\\) value there exists a corresponding \\(y\\) value that forms a solution pair.

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To solve a system that involves a linear equation and a non-linear equation we must use the substitution method.

\n

\\(\\var{a1}x+\\var{b1}y=\\var{r1}\\)

\n

\\(\\var{c1}x^2+\\var{d1}y^2=\\var{r2}\\)

\n

The first equation is a linear equaton, we use this to write one variable in terms of the other. 

\n

For example, we could make y the subject of this equation 

\n

\\(y=\\frac{\\var{r1}-\\var{a1}x}{\\var{b1}}\\)

\n

We can then insert this for every \\(y\\) in the non-linear equation to get

\n

\\(\\var{c1}x^2+\\var{d1}*\\left(\\frac{\\var{r1}-\\var{a1}x}{\\var{b1}}\\right)^2=\\var{r2}\\)

\n

\\(\\var{c1}x^2+\\var{d1}*\\frac{(\\var{r1}-\\var{a1}x)^2}{\\var{b1}^2}=\\var{r2}\\)

\n

Multiplying across by \\(\\simplify{{b1}^2}\\) gives

\n

\\(\\simplify{{c1}*{b1}^2}x^2+\\var{d1}(\\var{r1}-\\var{a1}x)^2-\\simplify{{r2}*{b1}^2}=0\\)

\n

\\(\\simplify{{c1}*{b1}^2}x^2+\\var{d1}\\left(\\simplify{{r1}^2}-\\simplify{2*{r1}*{a1}}x+\\simplify{{a1}^2}x^2\\right)-\\simplify{{r2}*{b1}^2}=0\\)

\n

Gathering the like terms together gives

\n

\\(\\simplify{({c1}*{b1}^2+{d1}*{a1}^2)x^2-(2*{a1}*{d1}*{r1})x+({d1}*{r1}^2-{r2}*{b1}^2)}=0\\)

\n

This is a quadratic equation.

\n

Recall the formula for solving a quadratic equation of the form  \\(ax^2+bx+c=0\\)  is given by

\n

\\(x=\\frac{-b\\pm \\sqrt{b^2-4ac}}{2a}\\)

\n

In this example \\(a = \\simplify{({c1}*{b1}^2+{d1}*{a1}^2)}, b = \\simplify{-(2*{a1}*{d1}*{r1})}\\) and \\(c = \\simplify{{d1}*{r1}^2-{r2}*{b1}^2}\\)

\n

Once we have each \\(x\\) value we insert it into   \\(y=\\frac{\\var{r1}-\\var{a1}x}{\\var{b1}}\\)   to find the corresponding \\(y\\) value.

", "parts": [{"showCorrectAnswer": true, "prompt": "

Input the larger of the two \\(x\\) values.      \\(x = \\) [[0]]

\n

Input the \\(y\\) value that corresponds to the previous answer.     \\(y = \\) [[1]]

\n

\n

Input the lesser of the two \\(x\\) values that satisfies both equations.     \\(x = \\) [[2]]

\n

Input the \\(y\\) value that corresponds to the previous answer.     \\(y = \\) [[3]]

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You should try every question!

"}, "preventleave": true, "startpassword": "3.1415"}, "timing": {"allowPause": true, "timeout": {"action": "warn", "message": "

Times up!

"}, "timedwarning": {"action": "warn", "message": "

You are going to time out soon...

"}}, "feedback": {"showactualmark": true, "showtotalmark": true, "showanswerstate": true, "allowrevealanswer": true, "advicethreshold": 0, "intro": "

This is a trial quiz to see how this program works.

", "reviewshowscore": true, "reviewshowfeedback": true, "reviewshowexpectedanswer": true, "reviewshowadvice": true, "feedbackmessages": [{"message": "

Great work!

", "threshold": "90"}, {"message": "

More practise needed!

", "threshold": "40"}], "enterreviewmodeimmediately": true, "showexpectedanswerswhen": "inreview", "showpartfeedbackmessageswhen": "always", "showactualmarkwhen": "always", "showtotalmarkwhen": "always", "showanswerstatewhen": "always", "showadvicewhen": "inreview"}, "type": "exam", "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}, {"name": "Jill Singleton", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2473/"}], "extensions": [], "custom_part_types": [], "resources": []}